What is the mass of a 202.6 mL sample of hydrogen chloride gas at -12.4^{o}C and 184.26 kPa

2,,,What volume of water vapor is created by the complete combustion of 35.5 liters of methane gas?

1. PV=nRT

V=202.6 ml, T=-12.4 C=273-12.4 K=260.6 K, P=184.26 kPa, R=8.314x10^3 cm3kPa K−1 mol−1

n=PV/RT=184.26x202.6/8.314x10^3x260.6=0.0172 mol

So the mas of HCl=0.0172x36.46=0.628 g

2. PV=nRT

Assume constant temperature and pressure, so the volume of the gas is proportional to the mole of the gas.

When CH4 is completely reacted,

CH4+O2 ==== CO2 + 2H2O

So the molar ration between H2O and CH4 is 2:1, hence the volume ratio between H2O and CH4 is 2:1.

Therefore 35.5 liters of methane gas could produce 2x35.5=70.0 liters of water vapor.

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