96213 mL = 0.096213 m3
3 atm = 303975 Pa
30 deg C = 303.15 K
The equation of the reaction (combustion) between benzaldehyde and oxygen is following:
C6H5CHO + 8 O2 = 7 CO2 + 3 H2O
The amount of benzaldehyde:
"n(C_6H_5CHO) = \\frac {m(C_6H_5CHO)}{M(C_6H_5CHO)} = \\frac{3025 g}{106 \\frac{g}{mol}} = 28.54 mol"
The amount of oxygen can be found using ideal gas law pV=nRT:
"n(O_2) = \\frac{pV}{RT} = \\frac {303975 Pa \\cdot 0.096213 m^3}{8.314 J\\cdot mol^{-1} \\cdot K^{-1} \\cdot 303.15 K} = 11.60 mol"
Looking at reaction stoichiometry one can conclude that benzaldehyde is in excess compared to oxygen. Calculating the amount of benzaldehyde reacted with given amount of oxygen:
"n(react. C_6H_5CHO) = \\frac {1}{8} n(O_2) = \\frac {1}{8} \\cdot 11.60 mol = 1.45 mol"
The amount of CO2 formed is:
"n(CO_2) = 7n(react.C_6H_5CHO) = 7 \\cdot 1.45 mol = 10.15 mol"
The amount of H2O formed is:
"n(H_2O) = 3n(react.C_6H_5CHO) = 3 \\cdot 1.45 mol = 4.35 mol"
The remnant of benzaldehyde is:
"n(remn. C_6H_5CHO) = n(C_6H_5CHO) - n(react.C_6H_5CHO) = 28.54 mol - 1.45 mol = 27.09 mol"
From three products of the reaction only CO2 is a gas at STP, thus calculating volume of CO2 using ideal gas law:
"V(CO_2) = \\frac {nRT}{p} = \\frac {10.15 mol \\cdot 8.314 J \\cdot mol^{-1} \\cdot K^{-1} \\cdot 273.15 K}{101325 Pa} = 0.227 m^3 = 227 L"
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