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CALCULATIONS INVOLVING LIMITING REAGENTS Sometimes in order to solve stoichiometric problems, you will need to find the limiting reagent first.

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Presentation on theme: "CALCULATIONS INVOLVING LIMITING REAGENTS Sometimes in order to solve stoichiometric problems, you will need to find the limiting reagent first."— Presentation transcript:

1 CALCULATIONS INVOLVING LIMITING REAGENTS Sometimes in order to solve stoichiometric problems, you will need to find the limiting reagent first.

2 Terms to Know Stoichiometry Stoichiometric coefficients Stoichiometric amounts Limiting reactant Excess reactant

3 Review: The limiting reactant gets used up first in a reaction. The excess reactant has some left over.

4 What is a Limiting Reactant? When you run out of one reactant before all the others. This will determine the maximum amount of product that can be produced. Think back to our miners complete breakfast…you have 12 slices of bread and 8 eggs,, 2000mL of juice and 17 slices of bacon - how many breakfasts can you make? You run out of eggs before you run out of anything else so you can only make 4 breakfasts’

5 The Limiting Reactant in Stoichiometric Problems After you determine the limiting reactant you use the mole value for the limiting reactant for all problems! E.g. N 2(g) + 3H 2(g)  2NH 3(g) m=25.0gm=5.00gm=? M=28.0g/molM=2.00g/mol n=m/M=25/28n=m/M=5/2 n=0.893moln=2.50mol Which one do you run out of first? Which is limiting?

6 Solution N 2(g) + 3H 2(g)  2NH 3(g) m=25gm=5gm=? M=28g/molM=2g/mol n=m/M=25/28n=m/M=5/2 n=0.893moln=2.5mol Compare N 2 to H 2 If you have 0.893 moles of N 2, how many moles of H 2 needed? 3H 2 /1N 2 = x H 2 /0.893 N 2 x = (3H 2 )(0.893N 2 )/1N 2 x = 2.68 moles of H 2 needed H 2 needed > H 2 actual, therefore H 2 is limiting excesslimiting n=0.893 molesn=2.5 moles calculate moles NH 3 produced 2NH 3 /3H 2 = x NH 3 /2.5H 2 x NH 3 = (2NH 3 )(2.5H 2 )/3H 2 x = 1.67 moles of NH 3 produced n = 1.67 moles M = 17 g/mole m = nM=(1.67)(17) m = 28.3 g of NH 3 formed

7 EXAMPLE 1: Determine the amount of titanium metal produced when 2.8 mol of titanium (IV) chloride reacts with 5.4 mol of magnesium. STEP 1: Write a balanced chemical equation with knowns and unknowns.

8 EXAMPLE 1: Determine the amount of titanium metal produced when 2.8 mol of titanium (IV) chloride reacts with 5.4 mol of magnesium. STEP 2: Find out which reactant is the limiting reagent by using one reactant to solve for the other or mole ratios.

9 EXAMPLE 1: Determine the amount of titanium metal produced when 2.8 mol of titanium (IV) chloride reacts with 5.4 mol of magnesium. STEP 3: Use the limiting reagent to find the number of moles of the required substance.

10 Now, we will follow the same logic to find the mass of a product. We always start by finding the limiting reagent.

11 EX 2: Methanol, CH3OH, is made by combining carbon monoxide and oxygen. What mass of CH3OH is produced from 9.80g of CO and 1.30g of H2? STEP 1: Write out the balanced chemical equation as well as knowns and unknowns underneath.

12 EX 2: Methanol, CH3OH, is made by combining carbon monoxide and oxygen. What mass of CH3OH is produced from 9.80g of CO and 1.30g of H2? STEP 2: Convert the mass of both substances to moles.

13 EX 2: Methanol, CH3OH, is made by combining carbon monoxide and oxygen. What mass of CH3OH is produced from 9.80g of CO and 1.30g of H2? STEP 3: Determine the limiting reagent.

14 EX 2: Methanol, CH3OH, is made by combining carbon monoxide and oxygen. What mass of CH3OH is produced from 9.80g of CO and 1.30g of H2? STEP 3: Use the limiting reagent to complete the question. Basically you just have to fill in the table.

15 Note: From now on, you must verify that you are using the limiting reactant when you are finding out how much product is formed.

16 HOMEWORK:


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