1.05 atm is 106391.25 Pa
35 deg. C is 308 K
Number of moles of H2 gas involved in the reaction:
"n(H_2) = \\frac{18.6 g}{2 \\frac{g}{mol}} = 9.3 mol"
According to the reaction equation quantity in moles of produced acetic acid is two times less than H2. Thus,
"n(CH_3COOH) = \\frac{9.3 mol}{2} = 4.65 mol"
Using general gas equation pV=nRT (where R is ideal gas constant) we are able to find the volume of acetic acid gas:
"V = \\frac{4.65 mol \\cdot 8.314 \\frac{m^3 \\cdot Pa}{K \\cdot mol} \\cdot 308 K}{106391.25 Pa} = 0.112 m^3 = 112 L"
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