Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 3 - Chemical Equations and Reaction of Stoichiometry - Exercises - Calculations Based on Chemical Equations - Page 107: 17


(a) $CaCO_3 + 2HCl -- \gt CaCl_2 + CO_2 + H_2O$ (b) $ 4.52 \times 10^{23} atoms$ (c) $9.03 \times 10^{22}molecules(CO_2)$

Work Step by Step

(a) 1. Begin by balancing the elements that appear only once in each side: - Balance the $Cl$: The subscript for chlorine in the products is "2", from $CaCl_2$. Therefore, we should put a "2" as the coefficient of $HCl$ $CaCO_3 + 2HCl -- \gt CaCl_2 + CO_2 + H_2O$ - The equation is balanced. (b) Each calcium carbonate molecules $CaCO_3$ has: 1 Calcium, 1 carbon and 3 oxygens, which gives us a total of 5 atoms. Use that as a conversion factor: $0.150mol(CaCO_3) \times \frac{5atoms}{1(CaCO_3)} = 0.750mol-atoms$ Convert that number to atoms: (1 mol = $6.022 \times 10^{23}$) $0.750 mol-atoms \times \frac{6.022 \times 10^{23}}{1 mol} = 4.52 \times 10^{23} atoms$ (c) According to the balanced equation, each mole of calcium carbonate produces 1 mole of carbon dioxide ($CO_2$). Use that as a conversion factor: $0.150mol(CaCO_3) \times \frac{1mol(CO_2)}{1mol(CaCO_3)} = 0.150mol(CO_2)$ Convert that number to molecules: (1 mol = $6.022 \times 10^{23}molecules$) $0.150mol(CO_2) \times \frac{6.022 \times 10^{23} molecules}{1mol} = 9.03 \times 10^{22}molecules(CO_2)$
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