General Chemistry 10th Edition

Published by Cengage Learning
ISBN 10: 1-28505-137-8
ISBN 13: 978-1-28505-137-6

Chapter 3 - Calculations with Chemical Formulas and Equations - Questions and Problems - Page 120: 3.41

Answer

a. 2.86 g C is 0.238 moles C. b. 7.05 g $Cl_{2}$ is 0.0994 mol $Cl_{2}$ c. 76 g $C_{4}H_{10}$ is 1.309 mol $C_{4}H_{10}$ d. 0.112 mol $Al_{2}(CO_{3})_3$

Work Step by Step

Strategy: We have to convert mass to moles. So the conversion factor is 1 mol/Molar mass. a. Moles in 2.86 g C. Molar mass of C is 12.01 g/mol Therefore : 2.86 g C x $\frac{1 mol C }{ 12.01 g}$ = 0.238 mol C (to three significant figures). So 2.86 g C is0.238 moles C. b. Moles in 7.05 g Cl2. Molar mass of $Cl_{2}$ = 2 x 35.45 Molar mass of $Cl_{2}$ = 70.9 g/mol. theerfore: 7.02 g x $\frac{1 mol Cl_{2}}{70.9 g}$ = 0.0994 mol $Cl_{2}$ So 7.05 g $Cl_{2}$ is 0.0994 mol $Cl_{2}$. c. 76 g $C_{4}H_{10}$ Molar mass $C_{4}H_{10}$ = 4 x 12.01 + 10 x 1.00 Molar mass $C_{4}H_{10}$ = 58.04 g/mol Therefore : 76g $C_{4}H_{10}$ x $\frac{1 mol_{ C_{4}H_{10}} }{ 58.04 g}$ = 1.309 mol So 76 g $C_{4}H_{10}$ is 1.309 mol $C_{4}H_{10}$ d. Moles in 26.2 g $Al_{2}(CO_{3})_3$ Molar mass $Al_{2}(CO_{3})_3$ =2 x 26.98 + 3 x12.01 + 9 x 15.99 Molar mass $Al_{2}(CO_{3})_3$ = 233.9 g/mol therefore: 26.2 g x $\frac{1 mol}{233.9 g}$ = 0.112 mol $Al_{2}(CO_{3})_3$ So 26.2 g $Al_{2}(CO_{3})_3$ s 0.112 mol $Al_{2}(CO_{3})_3$.
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