Limiting Reagent Problems Video Example. Limiting Reagent Problems Video Example Reagent = Reactant.

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Presentation transcript:

Limiting Reagent Problems Video Example

Limiting Reagent Problems Video Example Reagent = Reactant

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation

a)Determine the excess and limiting reagents.

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. b)What mass of AlBr 3 can be produced?

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. b)What mass of AlBr 3 can be produced? c)What mass of the excess reagent is left over?

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially.

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially.

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially.

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially.

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially.

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially.

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially.

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially.

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially.

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 1– Determine the number of moles of each reactant present initially mol Al present 2.50 mol Br 2 present

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts mol Al present 2.50 mol Br 2 present

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts mol Al present 2.50 mol Br 2 present

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts mol Al present 2.50 mol Br 2 present

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts mol Al present 2.50 mol Br 2 present

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts mol Al present 2.50 mol Br 2 present

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts mol Al present 2.50 mol Br 2 present

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts mol Al present 2.50 mol Br 2 present 2.78 mol Br 2 required

2.78 mol Br 2 required 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts mol Al present 2.50 mol Br 2 present 2.78 mol Br 2 required

2.78 mol Br 2 required 2.50 mol Br 2 present 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts mol Al present 2.50 mol Br 2 present 2.78 mol Br 2 required

2.78 mol Br 2 required 2.50 mol Br 2 present So Br 2 is the limiting reagent g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts mol Al present 2.50 mol Br 2 present 2.78 mol Br 2 required

2.78 mol Br 2 required 2.50 mol Br 2 present So Br 2 is the limiting reagent. And the Al is the excess reagent g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. Step 2– Determine the moles of Br 2 required to react with all the Al present, assuming all the Al reacts mol Al present 2.50 mol Br 2 present 2.78 mol Br 2 required

2.78 mol Br 2 required 2.50 mol Br 2 present So Br 2 is the limiting reagent. And the Al is the excess reagent g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents mol Al present 2.50 mol Br 2 present The moles of substances used up or produced must be based on the moles of the limiting reagent, Br 2, present

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Determine the excess and limiting reagents. b)What mass of AlBr 3 can be produced? c)What mass of the excess reagent is left over?

b)What mass of AlBr 3 can be produced? c)What mass of the excess reagent is left over? Initial moles Change in moles Final moles An ICF chart

b)What mass of AlBr 3 can be produced? c)What mass of the excess reagent is left over? Initial moles Change in moles Final moles

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation How many moles of AlBr 3 can be produced? 1.85 mol Al present 2.50 mol Br 2 present

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 3– Determine the moles of AlBr 3 produced mol Al present 2.50 mol Br 2 present How many moles of AlBr 3 can be produced?

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 3– Determine the moles of AlBr 3 produced mol Al present 2.50 mol Br 2 present The moles of substances used up or produced must be based on the moles of the limiting reagent, Br 2, present How many moles of AlBr 3 can be produced?

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 3– Determine the moles of AlBr 3 produced mol Al present 2.50 mol Br 2 present The moles of substances used up or produced must be based on the moles of the limiting reagent, Br 2, present How many moles of AlBr 3 can be produced?

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 3– Determine the moles of AlBr 3 produced mol Al present 2.50 mol Br 2 present Limiting Reagent How many moles of AlBr 3 can be produced?

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 3– Determine the moles of AlBr 3 produced mol Al present 2.50 mol Br 2 present How many moles of AlBr 3 can be produced?

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 3– Determine the moles of AlBr 3 produced mol Al present 2.50 mol Br 2 present How many moles of AlBr 3 can be produced?

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 3– Determine the moles of AlBr 3 produced mol Al present 2.50 mole Br 2 present How many moles of AlBr 3 can be produced?

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 3– Determine the moles of AlBr 3 produced mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced How many moles of AlBr 3 can be produced?

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation How many moles of Al are used up? Step 4– Determine the moles of Al used up mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 4– Determine the moles of Al used up mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Limiting Reagent How many moles of Al are used up?

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 4– Determine the moles of Al used up mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Limiting Reagent How many moles of Al are used up?

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 4– Determine the moles of Al used up mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced How many moles of Al are used up?

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 4– Determine the moles of Al used up mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced How many moles of Al are used up?

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation b)What mass of AlBr 3 can be produced? Step 4– Determine the moles of Al used up mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced

1.67 mol Al used up 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 4– Determine the moles of Al used up mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced How many moles of Al are used up?

1.67 mol Al used up 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles Change in moles Final moles

1.67 mol Al used up 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles Change in moles Final moles Excess Reagent

1.67 mol Al used up 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles Change in moles Final moles Excess Reagent Limiting Reagent

1.67 mol Al used up 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles1.85 Change in moles Final moles Excess Reagent Limiting Reagent

1.67 mol Al used up 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles Change in moles Final moles Excess Reagent Limiting Reagent

1.67 mol Al used up 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles Change in moles Final moles Excess Reagent Limiting Reagent

1.67 mol Al used up 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles Change in moles –1.67 Final moles Excess Reagent Limiting Reagent

1.67 mol Al used up 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles Change in moles –1.67–2.50 Final moles Excess Reagent Limiting Reagent

1.67 mol Al used up 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles Change in moles –1.67– Final moles

1.67 mol Al used up 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles Change in moles –1.67– Final moles0.18

1.67 mol Al used up 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles Change in moles –1.67– Final moles0.180

1.67 mol Al used up 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation Step 5– Produce an ICF chart for the three substances mol Al present 2.50 mol Br 2 present 1.67 mol AlBr 3 produced Initial moles Change in moles –1.67– Final moles

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation b)What mass of AlBr 3 can be produced? Step 6– Calculate the mass of AlBr 3 produced.

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation b)What mass of AlBr 3 can be produced? Step 6– Calculate the mass of AlBr 3 produced.

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation b)What mass of AlBr 3 can be produced? Step 6– Calculate the mass of AlBr 3 produced.

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation b)What mass of AlBr 3 can be produced? Step 6– Calculate the mass of AlBr 3 produced.

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation b)What mass of AlBr 3 can be produced? Step 6– Calculate the mass of AlBr 3 produced.

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation b)What mass of AlBr 3 can be produced? Step 6– Calculate the mass of AlBr 3 produced.

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation b)What mass of AlBr 3 can be produced? Step 6– Calculate the mass of AlBr 3 produced. 445 g of AlBr 3 is produced

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation c)What mass of the excess reagent is left over? 445 g of AlBr 3 is produced

50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation c)What mass of the excess reagent is left over? Step 7– Calculate the mass of Al left over.

445 g of AlBr 3 is produced 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation c)What mass of the excess reagent is left over? Step 7– Calculate the mass of Al left over.

445 g of AlBr 3 is produced 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation c)What mass of the excess reagent is left over? Step 7– Calculate the mass of Al left over.

445 g of AlBr 3 is produced 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation c)What mass of the excess reagent is left over? Step 7– Calculate the mass of Al left over.

445 g of AlBr 3 is produced 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation c)What mass of the excess reagent is left over? Step 7– Calculate the mass of Al left over.

445 g of AlBr 3 is produced 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation c)What mass of the excess reagent is left over? Step 7– Calculate the mass of Al left over.

445 g of AlBr 3 is produced 4.86 g of the excess reagent, Al, is left over 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation c)What mass of the excess reagent is left over? Step 7– Calculate the mass of Al left over.

445 g of AlBr 3 is produced 4.86 g of the excess reagent, Al, is left over When 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation

445 g of AlBr 3 is produced 4.86 g of the excess reagent, Al, is left over When 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Al is the excess reagent

445 g of AlBr 3 is produced 4.86 g of the excess reagent, Al, is left over When 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Al is the excess reagent and Br 2 is the limiting reagent.

445 g of AlBr 3 is produced 4.86 g of the excess reagent, Al, is left over When 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Al is the excess reagent and Br 2 is the limiting reagent. b)445 g of AlBr 3 is produced.

445 g of AlBr 3 is produced 4.86 g of the excess reagent, Al, is left over When 50.0 g of Al (s) is mixed with g of Br 2(l) and allowed to react according to the balanced equation a)Al is the excess reagent and Br 2 is the limiting reagent. b)445 g of AlBr 3 is produced. c)4.86 g of Al, the excess reagent, is left over after the reaction.