# Question #7cee0

You don't write the reaction (that must be balanced) but every decomposition of 1 mole of $M g C {O}_{3}$ normally give 1 mole of carbon dioxide.
Since 1 mol of gas at STP (pressure and temperature standard, normally 1 Atm and 273°C ) occupy 22,4 L, to have 661 L of gas you must have $\frac{661 L}{22 , 4 \frac{L}{m} o l} = 29 , 5 m o l$ of $C {O}_{2}$ that came from 29,5 mol of $M g C {O}_{3}$ that are
$29 , 5 m o l 84 , 3 \frac{g}{m o l} = 2488 g$