10 gmol 164 g CaNO32 b 10 x 10 2 mol CaOH2 x 2 mol HNO31 mol CaOH2 2 x 10 2 mol

# 10 gmol 164 g cano32 b 10 x 10 2 mol caoh2 x 2 mol

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mol x 164.10 g/mol = 1.64 g Ca(NO3)2 b) 1.0 x 10 -2 mol Ca(OH)2 x 2 mol HNO3/1 mol Ca(OH)2 = 2 x 10 -2 mol of Ca(OH)2 What is left over: 3.00 x 10 -2 mol 1.0 x 10 -2 mol = 2.0 x 10 -2 mol HNO3 2. Consider the following reaction: P (s) + S (s) P 4 S 3 (s) a) How many grams of tetraphosphorus trisulfide can be produced from 62.0 g of phosphorus and 38.6 g of sulfur? B) How many grams of which reagent will remain unreacted? 4 P (s) + 3 S (s) P 4 S 3 (s) a) n = m/M = 62.0 g/30.97 g/mol = 2.00 mol P n = m/M = 38.6 g/32.07 g/mol = 1.20 mol S 2.00 mol P x 3 mol S/4 mol P = 1.50 mol S : The limiting reagent is S . 1.20 mol S x 1 mol P4S3/3 mol S = 0.400 mol P4S8 m = nM = 0.400 mol x 220.06 g/mol = 88.0 g of P4S3 b) Excess reagent: 1.2 mol S x 4 mol P/3 mol S = 1.6 mol P will be used up Unreacted reagent: 2.00 mol 1.6 mol = 0.400 mol of P will remain . M = nM = 0.400 mol x 30.97 g/mol = 12.388 g of P will remain .
1.11.1 Exercises: 3. Consider the following reaction: HBr (aq) + Fe(OH) 3 (s) FeBr 3 (aq) + H 2 O (l) a) What weight of iron (III) bromide can be produced by the reaction of 30.0 g of hydrogen bromide with 10.0 g of iron (III) hydroxide? B) Given that the percentage yield is 66%, how many grams of iron (III) bromide are obtained? 3 HBr (aq) + Fe(OH) 3 (s) FeBr 3 (aq) + 3 H 2 O (l) a) HBr: n = m/M = 30.0 g/80.91 g/mol = 0.371 mol HBr Fe(OH)3: n = m/M = 10.0 g/106.88 g/mol = 0.0936 mol Fe(OH)3 Limiting reagent calculations: 0.371 mol HBr x 1 mol Fe(OH)3/3 mol HBr = 0.124 mol Fe(OH)3 Fe(OH)3 is the limiting reagent. 0.0936 mol Fe(OH)3 x 1 mol FeBr3/1 mol Fe(OH)3 = 0.0936 mol m = nM = 0.0936 mol x 295.55 g/mol = 27.66 g FeBr3 b) 66% x 27.66 g = 18.26 g
1.12.1 Exercises: 1. A solution can be prepared by mixing 1.00 g of ethanol (C 2 H 5 OH) with 100.0 g of water to give a final volume of 101 mL. Calculate the molarity of ethanol in the solution. n = m/M = 1.00 g/46.08 g/mol = 0.0217 mol M = n/v = 0.0217 mol/0.101 l = 0.215 mol/L 2. A sulphuric acid solution of density 1.802 g/mL contains 88.0% H 2 SO 4 by weight. A) Find the weight of H 2 SO 4 per liter of solution. B) Find the molarity of the solution. a) 1 liter = 1 802 g 0.880 x 1 802 g = 1.59 x 10 3 g b) n = m/M = 1.59 x 10 3 g/98.09 g/mol = 16.2 mol M = n/v = 16.2 mol/1 L = 16.2 mol/L 3. A 325 mL sample of a solution contains 25.3 g CaCl 2 (111 g/mol). A) Calculate the molarity of Cl - ions in solution. B) How many grams of Cl - are there in 0.100L of this solution? a) n = m/M = 25.3 g/110.98 g/mol = 0.2280 mol M = n/v = 0.2280 mol/0.325 L = 0.702 mol/L b) n = MV = 0.702 mol/L x 0.100 L = 0.0702 mol m = nM = 0.0702 mol x 35.45 g/mol = 2.49 g of Cl - . 4. Find the weight percent of calcium carbonate in a 2.00 g sample of rock if it reacts with 30.0 mL of 0.60 M H 2 SO 4 according to the following reaction: CaCO 3 (s) + H 2 SO 4 (aq) CaSO 4 (s) + H 2 O (l) + CO 2 (g) n = MV = 0.60 mol/L x 0.0300 L = 0.0180 mol m = nM = 0.0180 mol x 100.09 g/mol = 1.80 g Weight % = 1.80 g/2.00 g x 100% = 90.0%

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