Pressure = 1.12atm
Temperature = 40°C + 273 = 313K
"20g H_2O \u00d7\\frac{1 mole H_2}{1 mole H_2O}=1.1099molH_2"
"V=\\frac{nRT}{P}"
"V=(1.1099)(\\frac{0.112Latm}{mol\u00d7K})(313K)\u00f7112atm =34.74LH_2"
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