Question

The reaction involved in the explosive combusion of acetylene is : 2 C2H2 (g) + 5 O2(g) g 4 CO2(g) + 2 H2 (g) How many L of CO2 gas (measured at STP) will be formed during the combustion of 50.0 L C2H2 gas (measured at STP)

Answer 1

2C₂H₂(g) + 5O₂(g) g → 4CO₂(g) + 2H₂(g)

At STP one mole of any gas occupies 22.4 L

n(C₂H₂) $= \frac{50.0}{22.4} = 2.232 \;mol$

According to the reaction:

n(CO₂) = 2n(C₂H₂) $= 2 \times 2.232 = 4.464 \;mol$

V(CO₂) $= 4.464 \times 22.4 = 99.99 ≈ 100 \;L$

Answer: 100 L