Question #7cee0

1 Answer
Apr 20, 2017

2488 g

Explanation:

You don't write the reaction (that must be balanced) but every decomposition of 1 mole of #MgCO_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 # (661L)/(22,4 L/mol) = 29,5 mol# of #CO_2# that came from 29,5 mol of #MgCO_3# that are
#29,5 mol 84,3 g/(mol)= 2488 g #