University:
Felician University
Course:
CHEM-105 | Principles of Chemistry
Academic year:

2020
Views:

134

Pages:

2
Author:

Clint W.

**Stoichiometry Worksheet** 1. Na2SiO3(s) + 8 HF(aq) → H2SiF6(aq) + 2 NaF(aq) + 3 H2O(l) a. How many moles of HF are needed to react with 0.300 mol of Na2SiO3? 0.300 mol Na2SiO3 * (8 mol HF / 1 mol Na2SiO3) = 2.40 mol HF b. How many grams of NaF form when 0.500 mol of HF reacts with excess Na2SiO3? 0.500 mol HF * (2 mol NaF / 8 mol HF) * (41.99 g NaF / 1 mol NaF) = 5.25 g NaF c. How many grams of Na2SiO3 can react with 0.800 g of HF? 0.800 g HF * (1 mol HF / 20.01 g HF) * (1 mol Na2SiO3 / 8 mol HF) * (122.06 g Na2SiO3 / 1 mol Na2SiO3) = 0.613 g Na2SiO3 2. C6H12O6(aq) → 2 C2H5OH(aq) + 2 CO2(g) a. How many moles of CO2 are produced when 0.400 mol of C6H12O6 reacts in this fashion? 0.400 mol C6H12O6 * (2 mol CO2 / 1 mol 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 / 46.07 g C2H5OH) * (1 mol C6H12O6 / 2 mol C2H5OH) * (180.18 g C6H12O6 / 1 mol C6H12O6) = 14.7 g C6H12O6 c. How many grams of CO2 form when 7.50 g of C2H5OH are produced? 7.50 g C2H5OH * (1 mol C2H5OH / 46.07 g C2H5OH) * (2 mol CO2 / 2 mol C2H5OH) * (44.01 g CO2 / 1 mol CO2) = 7.16 g CO2 3. Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) (unbalanced!) a. Calculate the number of grams of CO that can react with 0.150 kg of Fe2O3. 0.150 kg Fe2O3 * (1000 g Fe2O3 / 1 kg) * (1 mol Fe2O3 / 159.69 g Fe2O3) * (3 mol CO / 1 mol Fe2O3) * (28.01 g CO / 1 mol CO) = 78.9 g CO b. Calculate the number of grams of Fe and the number of grams of CO2 formed when 0.150 kg of Fe2O3 reacts. 150g Fe2O3 * (1 mol Fe2O3 / 159.69 g Fe2O3) * (2 mol Fe / 1 mol Fe2O3) * (55.85 g Fe / 1 mol Fe) = 105 g Fe 150g Fe2O3 * (1 mol Fe2O3 / 159.69 g Fe2O3) * (3 mol CO2 / 1 mol Fe2O3) * (44.01 g CO2 / 1 mol CO2) = 124 g 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 CO2 are allowed to react? 1.85 mol NaOH * (1 mol CO2 / 2 mol NaOH) = 0.925 mol CO2 1.00 mol CO2 * (1 mol CO2 / 1 mol CO2) = 1.00 mol CO2 L.R. = NaOH b. How many moles of Na2CO3 can be produced? 1.85 mol NaOH * (1 mol Na2CO3 / 2 mol NaOH) = 0.925 mol Na2CO3 **5. C6H6 + Br2 → C6H5Br + HBr** a. What is the theoretical yield of C6H5Br in this reaction when 30.0 g of C6H6 reacts with 65.0 g Br2? 30.0 g C6H6 * (1 mol C6H6 / 78.11 g C6H6) = 0.384 mol 65.0 g Br2 * (1 mol Br2 / 159.81 g Br2) = 0.406 mol 0.384 mol C6H6 * (1 mol C6H5Br / 1 mol C6H6) * (157.01 g C6H5Br / 1 mol C6H5Br) = 60.3 g C6H5Br b. If the actual yield of C6H5Br was 56.7 g, what is the percent yield? (56.7 g / 60.3 g) * 100% = 94%