1. For the following reaction, 5.94 g of water are mixed with excess sulfur diox
ID: 589227 • Letter: 1
Question
1. For the following reaction, 5.94 g of water are mixed with excess sulfur dioxide. the reaction yield 21.3 g of sulfurous acid. what is the theoretical yoeld of sulfurous acid in grams? what is the percent yield for this reaction?2. For the following reaction 3.02 g of water are mixed with excess phosphorus pentoxide. The reaction yields 7.19 g of phosphoric acid. what is the theoretical yield of phosphoric acid in grams? what is the percent yield for this reaction?
3. For the following reaction 3.79 g of sulfur are mixed with excess carbon monoxide. the reaction yields 5.63 g of sulfur dioxide. what is the theoretical yield of sulfur dioxide? what is the percent yield of sulfur dioxide?
1. For the following reaction, 5.94 g of water are mixed with excess sulfur dioxide. the reaction yield 21.3 g of sulfurous acid. what is the theoretical yoeld of sulfurous acid in grams? what is the percent yield for this reaction?
2. For the following reaction 3.02 g of water are mixed with excess phosphorus pentoxide. The reaction yields 7.19 g of phosphoric acid. what is the theoretical yield of phosphoric acid in grams? what is the percent yield for this reaction?
3. For the following reaction 3.79 g of sulfur are mixed with excess carbon monoxide. the reaction yields 5.63 g of sulfur dioxide. what is the theoretical yield of sulfur dioxide? what is the percent yield of sulfur dioxide?
2. For the following reaction 3.02 g of water are mixed with excess phosphorus pentoxide. The reaction yields 7.19 g of phosphoric acid. what is the theoretical yield of phosphoric acid in grams? what is the percent yield for this reaction?
3. For the following reaction 3.79 g of sulfur are mixed with excess carbon monoxide. the reaction yields 5.63 g of sulfur dioxide. what is the theoretical yield of sulfur dioxide? what is the percent yield of sulfur dioxide?
Explanation / Answer
Molar mass of sulfur dioxide = 64.066 g/mol
Molar mass of water = 18.015 g/mol
Molar mass of sulfurous acid = 82.08 g/mol
Molar mass of phosphorus pentoxide= 283.886 g/mol
Molar mass of phosphoric acid = 97.994 g/mol
Molar mass of carbon monoxide = 28.01 g/mol
Molar mass of sulfur= 32.065 g/mol
Molar mass of sulfur dioxide = 64.066 g/mol
1.
SO2 + H2O --> H2SO3
Mass of water mixed = 5.94 g
Moles of water mixed = 5.95 g / 18.015 g/mol = 0.33
By stoichiometry, theoretical moles of sulfurous acid produced = 0.33 * 1/1 = 0.33
Theoretical yield of sulfurous acid produced = 0.33 mol *82.08 g/mol = 27.1 g
% yield of reaction = actual yield / theoretical yield * 100
= 21.3 / 27.1 * 100 = 78.57 %
2.
P4O10 + 6 H2O --> 4 H3PO4
Mass of water mixed = 3.02 g
Moles of water mixed = 3.02 g / 18.015 g/mol = 0.168
By stoichiometry, theoretical moles of phosphoric acid produced = 0.168 * 4/6 = 0.11
Theoretical yield of phosphoric acid produced = 0.11 mol *97.994 g/mol = 10.95 g
% yield of reaction = actual yield / theoretical yield * 100
= 7.19 / 10.95 * 100 = 65.65 %
3.
2 CO + S --> SO2 + 2 C
Mass of sulfur mixed = 3.79 g
Moles of sulfur mixed = 3.79 g / 32.065 g/mol = 0.118
By stoichiometry, theoretical moles of sulfur dioxide produced = 0.118 * 1/1 = 0.118
Theoretical yield of sulfur dioxide produced = 0.118 mol * 64.066 g/mol = 7.57 g
% yield of reaction = actual yield / theoretical yield * 100
= 5.63 / 7.57 * 100 = 74.35 %
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.