Question

3025 grams of benzaidehyde CHICHO) reacts with 96213 mL of Oxygen ges at a pressure of 3 atm and a amperabure ofl 30 degrees celsius. Deine the amount of products
Determine the volume of the products produced if the products are brought to STP

Answer 1

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:

$V(CO_2) = \frac {nRT}{p} = \frac {10.15 mol \cdot 8.314 J \cdot mol^{-1} \cdot K^{-1} \cdot 273.15 K}{101325 Pa} = 0.227 m^3 = 227 L$