Stoichiometry worksheet key PDF

Preview

Summary

Stoichiometry...

Description

Stoichiometry Worksheet 1. Na2SiO3(s) + 8 HF(aq)  H2SiF6(aq) + 2NaF(aq) + 3 H2O(l) a. How many moles of HF are needed to react with 0.300 mol of Na 2SiO3? 0.300 mol Na2Sio3 * 8mol HF/1mol Na2SiO3 = 2.4 mol HF b. How many grams of NaF form when 0.500 mol of HF reacts with excess Na 2SiO3? 0.500mol HF * 2mol NaF/8mol HF * 42.00 g/mol NaF = 5.25 g NaF c. How many grams of Na2SiO3 can react with 0.800 g of HF? 0.800g HF * 1 mol HF/20.00g HF * 1mol Na2SiO3/8 mol HF * 122.08 g Na2SiO3/1 mol = 0.610g Na2SiO3 2. C6H12O6(aq)  2C2H5OH(aq) + 2CO2(g) a. How many moles of carbon dioxide are produced when 0.400 mol of C 6H12O6 reacts in this fashion? 0.400mol C6H12O6 * 2mol CO2/1mol C6H12O6 = 0.800 mol CO2 b. How many grams of C6H12O6 are needed to form 7.50 g of C2H5OH? 7.50 g C2H5OH * 1 mol C2H5OH/46.00g * 1mol C6H12O6/2mol C2H5OH * 180.00g/1mol = 14.67 g C6H12O6 c. How many grams of CO2 form when 7.50 g of C2H5OH are produced? 7.50 g C2H5OH * 1 mol C2H5OH/46.00g * 2mol CO2/2mol C2H5OH * 44.01 g CO2/1 mol = 7.18g CO2 3. Fe2O3(s) + CO(g)  Fe(s) + CO2(g) (unbalanced) Fe2O3(s) +3 CO(g)  2Fe(s) + 3CO2(g) a. Calculate the number of grams of CO that can react with 0.150 kg of Fe 2O3. 0.150kg *1000g/kg * 1 mol/159.70g * 3 mol CO/1mol Fe2O3 * 28.01 g CO/mol = 78.9g b. Calculate the number of grams of Fe and the number of grams of CO 2 formed when 0.150 kg of Fe2O3 reacts. 0.150kg *1000g/kg * 1 mol/159.70g * 2mol Fe/1mol Fe2O3 *55.85g/mol = 104.9gFe

0.150kg *1000g/kg * 1 mol/159.70g * 3 mol CO2/1mol Fe2O3 * 44.01 g CO2/mol = 124g CO2

4. 2NaOH(s) + CO2(g)  Na2CO3(s) + H2O(l) a. Which reagent is the limiting reactant when 1.85 mol NaOH and 1.00 mol CO 2 are allowed to react? 1.85mol NaOH * 1mol Na2CO3/ 2mol NaOH = 0.925mol Na2CO3 Limiting reagent 1.00mol CO2 * 1mol Na2CO3/1mol CO2 = 1.00 mol Na2CO3

b. How many moles of Na2CO3 are produced? 1.85mol NaOH * 1mol Na2CO3/ 2mol NaOH = 0.925mol Na2CO3

5. C6H6(l) + Br2(l)  C6H5Br (l) + HBr(g) a. What is the theoretical yield of C6H5Br in this reaction when 30.0 g C6H6 reacts with 65.0 g of Br2? 30.0g C6H6 * 1mol/78.06g * 1mol C6H5Br/1mol C6H6 * 156.96g/1mol = 60.32 g Limiting reagent 65.0 g Br2 * 1mol/159.8g Br2 * 1mol C6H5Br/1mol Br2 * 156.96g/1mol = 63.84g

b. If the actual yield of C6H5Br was 56.7 g, what is the percent yield? 56.7g / 60.32g * 100 = 94.0% Yield...