## Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning

# Chapter 3 - Chemical Reactions - Questions for Review and Thought - Topical Questions - Page 149e: 65c

112 g

#### Work Step by Step

1. As we calculated on the last exercise, this reaction used 0.635 mol of $CCl_2F_2$, and the balanced reaction is: $1CCl_2F_2 + 2Na_2C_2O_4 -- \gt C + 4CO_2 + 2NaCl + 2NaF$ So, the ratio of $CCl_2F_2$ to $CO_2$ is 1 to 4: $0.635 mol (CCl_2F_2) \times \frac{4 mol(CO_2)}{1 mol (CCl_2F_2)} = 2.54mol (CO_2)$ 2. Calculate the mass of $CO_2$: Molar mass: 12.01* 1 + 16.00* 2 = 44.01g/mol $2.54 mol \times \frac{ 44.01 g}{ 1 mol} = 112g (CO_2)$

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