University:
Concordia University Wisconsin
Course:
CHEM 1414 | General Chemistry I
Academic year:

2021
Views:

469

Pages:

4
Author:

Prince Sampson

STOICHIOMETRY - a review I. CALCULATIONS INVOLVING MOLES, MASS AND VOLUME: In the reaction: 2C2H6 + 7O2 → 4CO2 + 6H2O a) How many moles of CO2 are produced when 18 moles of C2H6 are consumed? 18 C2H6 x (4 CO2/ 2 C2H6) = 36 moles b) How many moles of O2 are consumed when 13 moles of CO2 are produced? 13 CO2 x (7 O2 / 4 CO2) = 23 mol Hydrazine (N2H4) is a rocket fuel which is prepared according to the reaction: 2NH3 + NaOCl (bleach) → N2H4 + NaCl + H2O a) How many moles of hydrazine could be produced from 25.6 moles of bleach? 25.6 mol NaOCl x (1 N2H4 / 1 NaOCl) = 25.6 mol b) What mass (g) of hydrazine could be produced from 528.2 g of ammonia? 528.2 g NH3 x (1 mol / 17.0 g) x (1 N2H4 / 2 NH3) x (32.0 g / mol) = 497 g A mine produces a silver ore named argentite (Ag2S). The ore is then smelted according to the reaction shown to yield pure silver. What volume of CO2 (STP) would be produced when 800.00 g of silver is produced? Ag2S + C + 2O2 → 2Ag + CO2 + SO2 800.0 g Ag x (1 mol / 107.9 g) x (1 CO2 / 2 Ag) x (22.4 L / mol) = 83.0 L CO2 II. CALCULATIONS INVOLVING MOLARITY AND DILUTIONS How many g of salt (NaCl) would need to make 1.5 L of a 2.5 M solution? 1.5 L x (2.5 mol/L) x 58.5 g/mol = What volume of 0.10 M HCl would be required if a person needs to react 0.45 moles of HCl? V = mol / M = 0.45 mol / 0.1 mol/L = 0.045 L = 45 mL What volume of a 0.75 M solution sugar would be required if a reaction requires 0.25 moles of sugar? V = mol / M = 0.25 mol / 0.75 mol/L = 0.33 L or 3.3 x 10^2 mL Lulu needs to prepare 750.0 mL of a 2.5 M solution of sulfuric acid (H2SO4). a) How many mLs of pure (18.0 M) H2SO4 will she need? M1V1 = M2V2 18 V1 = 2.5 x 750 = 104 mL c) How much water will she need to add to the acid to make the solution? 750 - 104 = 646 mL Use code with caution. A 0.75 M lead II hydroxide is reacted with 50.0 mL of a 1.5 M solution of sodium iodide. Pb(OH)2 + 2NaI --> PbI2 + 2NaOH How many mL of lead II hydroxide will be required to used up all of the sodium iodide solution? 0.050 L x (1.5 mol / 1 L NaI) x (1 mol Pb(OH)2 / 2 mol NaI) = 0.0375 mol Pb(OH)2 V = mol / M = 0.0375 mol / 0.75 mol/L = 0.050 L or 50 ml