## Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.

# Chapter 3 - Chemical Equations and Reaction of Stoichiometry - Exercises - Calculations Based on Chemical Equations - Page 107: 14

#### Answer

(a) $CaCO_3 + 2HCl -- \gt CaCl_2 + CO_2 + H_2O$ (b) 5.2 mol $(HCl)$ (c) 2.6 mol $(H_2O)$

#### Work Step by Step

(a) - Write the equation with chemical formulas: $CaCO_3 + HCl -- \gt CaCl_2 + CO_2 + H_2O$ 1. Begin with the elements that appear exactly 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. ------------- Use the balance coefficients as conversion factors: (b) According to the balanced equation, each mole of $CaCO_3$ needs 2 moles of $HCl$ to react: $2.6 mol (CaCO_3) \times \frac{2mol(HCl)}{1mol(CaCO_3)} = 5.2 mol(HCl)$ (c) According to the balanced equation, each mole of $CaCO_3$ produces one mol of $H_2O$: $2.6 mol(CaCO_3) \times \frac{1mol(H_2O)}{1mol(CaCO_3)} = 2.6 mol(H_2O)$

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