# WE BE COOKING…. OH CRAP I ONLY HAVE ½ CUP OF SUGAR! Limiting Reagents.

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WE BE COOKING…. OH CRAP I ONLY HAVE ½ CUP OF SUGAR! Limiting Reagents

Brownies again… Only have ½ cup of sugar…bummer….what to do? ½ cup butter 2 oz of chocolate 1 cp sugar 2 eggs 1 tsp vanilla 2/3 cp flour ½ tsp baking powder ¼ tsp salt 24 Brownies Need to do some Stoiching to figure out how much of other ingredients if only have ½ cup sugar…. Sugar is the limiting reagent Everything else in excess.

Limiting Reagent Demo  Solid zinc is added to 6.00 M HCl  Zinc will all dissolve but….  pH is still below 7…meaning still have H + ions present Which reagent is limiting?  Zinc Which reagent is excess?  HCl

Stoichiometry with Limiting Reagent How to determine the limiting and excess reagent  1. Get both reagents into the unit moles  2. Divide those moles found by the coefficient of the reagent  3. The smaller number is the limiting reagent (LR)  4. The larger number is in excess  5. All stoich problems will be determined by the limiting reagent (LR)…therefore the given for DA will be the LR

Example with Limiting Reagent 7.24 moles of magnesium is added to 3.86 moles of oxygen gas to make MgO. Which reactant is used up? Excess? 2Mg(s) + O 2 (g)  2MgO(s) 7.24 mol Mg 2 = 3.62 mol of Mg 3.86 mol O 2 1 = 3.86 mol of O 2 Mg is LR O2 is excess 7.24 mol of Mg controls the reaction, therefore 7.24 mol will be your given for any DA

Example with Limiting Reagent 7.24 moles of magnesium is added to 3.86 moles of oxygen gas to make MgO. How many grams of MgO is produced? How many grams of oxygen is needed? 2Mg(s) + O 2 (g)  2MgO(s) 7.24 mol Mg 2 mol Mg 2 mol MgO 1 mol MgO 40.304 g MgO = 292 g of MgO 7.24 mol Mg 2 mol Mg 1 mol O 2 31.998 g O 2 = 116 g of O 2

Remember… Must work with the Limiting Reagent in Stoich  Must get both reagents into moles  Divide each by their coefficient  The smaller number is the LR…controls the reaction  This will be your given in DA

Problem If 7.56 gram of iron metal are placed in 100. mL of a 1.00 M hydrochloric acid, HCl, solution, hydrogen gas and iron(II) chloride are produced. Which reactant is limiting? How many grams are in excess? How many grams of each product is formed? 1. Balance chemical reaction Fe(s) + 2HCl(aq)  H 2 (g) + FeCl 2 (aq)

Problem If 7.56 gram of iron metal are placed in 100. mL of a 1.00 M hydrochloric acid, HCl, solution, hydrogen gas and iron(II) chloride are produced. Which reactant is limiting? How many grams are in excess? How many grams of each product is formed? Fe(s) + 2HCl(aq)  H 2 (g) + FeCl 2 (aq) 2. Find moles of each reactant to find LR 7.56 g Fe 55.845 g 1 mol Fe = 0.135 mol Fe 1 100. mL 1000 mL 1 L1.00 mol HCl 1 L = 0.100 mol HCl 2 = 0.0500 mol HCl LR

Problem If 7.56 gram of iron metal are placed in 100. mL of a 1.00 M hydrochloric acid, HCl, solution, hydrogen gas and iron(II) chloride are produced. Which reactant is limiting? How many grams are in excess? How many grams of each product is formed? Fe(s) + 2HCl(aq)  H 2 (g) + FeCl 2 (aq) 3. Find grams of excess (find grams you actually need) 0.100 mol HCl is LR 0.100 mol HCl 2 mol HCl 1 mol Fe 55.845 g Fe = 2.79 g of Fe 7.56 gram Fe – 2.79 gram Fe = 4.77 gram Fe excess

Problem If 7.56 gram of iron metal are placed in 100. mL of a 1.00 M hydrochloric acid, HCl, solution, hydrogen gas and iron(II) chloride are produced. Which reactant is limiting? How many grams are in excess? How many grams of each product is formed? Fe(s) + 2HCl(aq)  H 2 (g) + FeCl 2 (aq) 3. Find grams of each product (H 2 and FeCl 2 ) 0.100 mol HCl is LR 0.100 mol HCl 2 mol HCl 1 mol FeCl 2 126.745 g FeCl 2 = 6.34 g of FeCl 2 0.100 mol HCl 2 mol HCl 1 mol H 2 2.0158 g H 2 = 0.101 g of H 2

Review Steps for Stoiching… Write and balancing chemical reaction Find limiting reagent  Determine moles of each reactant  Divide the moles be their coefficient  The smaller number is LR Use the moles of the LR to convert to anything else in the problem  Moleland!! Convert to moles of another compound/atom Convert to final unit (moles/grams/volume)

Percent Yield 100% is maximum Percent yield measures how well you did the lab Compares lab measurement (actual) over theoretical amount (found using Stoich) % Yield = Actual amount (grams or moles) [Lab] Theoretical amount (grams or moles) [Stoich] Actual amount (grams or moles) x 100

If 12.5 grams of copper are reacted with an excess of chlorine gas, then 25.4 grams of copper(II) chloride, CuCl2(s), are obtained. Calculate theoretical amount and percent yield. 1 mol CuCl 2 Percent Yield 12.5 g Cu 63.546 g 1 mol 1 mol CuCl 2 134.446 g CuCl 2 Molar mass of CuCl 2 = 26.4 g of CuCl 2 Coefficients 1 mol Cu Cu(s) + Cl 2 (g)  CuCl 2 (s) Find grams CuCl 2 Molar mass of Cu % Yield = 25.4 g 26.4 g x 100= 96.4%

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