Answer
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
$2.54 mol \times \frac{ 44.01 g}{ 1 mol} = 112g (CO_2)$
