# Question #690f9

Jan 14, 2017

$1.12 {\mathrm{dm}}^{3}$
We have 5.4g of ${N}_{2} {O}_{5}$ which has an ${M}_{r}$ of 108 (14x2) + (16x5).
Moles of ${N}_{2} {O}_{5}$ = mass/${M}_{r}$ = 5.4/108 = 0.05 mol
The molar volume of a gas at STP = $22.4 {\mathrm{dm}}^{3}$
Thererfore the volume of 0.05 mol of the gas = 0.05 x 22.4 = $1.12 {\mathrm{dm}}^{3}$