Question

What volume of hydrogen gas at 35.0℃ and 106.6 kPa is needed to react completely with 120.0 g of tin to form tin (IV) hydride?

Answer 1

$2H_{2(g)}+Sn_{(s)}\to\>H_4Sn_{(s)}$

Molar mass of tin$=118.71$

Moles$=\frac{120}{118.71}=1.0109$

Moles of Hydrogen$=2×1.0109$

$=2.0218$

PV$=nRT$

$V=\frac{2.0218×8.3145×(273+35)}{106.6×10^3}$

$=0.04857m^3$