Answer
170 g
Work Step by Step
1. Calculate the number of moles of $CCl_2F_2$:
12.01* 1 + 35.45* 2 + 19.00* 2 = 120.91g/mol $(CCl_2F_2)$:
$76.8g \times \frac{1 mol}{120.91g} = 0.635mol (CCl_2F_2)$
2. As we determined on the last exercise, 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 $Na_2C_2O_4$ is 1 to 2:
$0.635 mol (CCl_2F_2) \times \frac{2 mol(Na_2C_2O_4)}{1 mol (CCl_2F_2)} = 1.27mol (Na_2C_2O_4)$
3. Calculate the mass of $Na_2C_2O_4$:
22.99* 2 + 12.01* 2 + 16* 4 = 134g/mol
$1.27 mol \times \frac{ 134 g}{ 1 mol} = 170g (Na_2C_2O_4)$
