= 560 L
1 mole of a gas at STP occupies a volume of 22400 cm³
Theerefore; in this question we find the volume occupied by 1.1x10^3 grams of CO2
1 mole of CO2 = 22.4 L
but, 1 mole of CO2 = 44 g
thus; 44 g = 22.4 L
Hence; 1.1x10^3 grams will occupy
= ((1.1x10^3 grams)/44)× 22.4