Step-by-step explanation:
A) This is a combustion reaction, so acetylene gas will react with oxygen gas to give carbon dioxide and water. The balanced reaction is:
C₂H₂ + 5/2 O₂ ----> 2 CO₂ + H₂O
2 C₂H₂ + 5 O₂ ----> 4 CO₂ + 2 H₂O
To solve this problem we will have to suppose that the gases are at Standard conditions of temperature and pressure (STP). One mol of any gas will occupy 22.4 L at STP. We can use that relationship to find the number of moles of CO₂ that we have in 75.0 L.
1 mol of CO₂ = 22.4 L
moles of CO₂ = 75.0 L * 1 mol/(22.4 L)
moles of CO₂ = 3.35 mol
2 C₂H₂ + 5 O₂ ----> 4 CO₂ + 2 H₂O
According to the coefficients of the reaction, 2 moles of C₂H₂ will react with 5 moles of O₂ to produce 4 moles of CO₂ and 2 moles of H₂O. So to produce 4 moles of CO₂ we will need 2 moles of C₂H₂. So the molar ratio between CO₂ and C₂H₂ is 4 to 2. We can use this ratio to find the number of moles of C₂H₂ necessary to produce 3.35 moles of CO₂.
4 moles of CO₂ = 2 moles of C₂H₂
moles of C₂H₂ = 3.35 moles of CO₂ * 2 moles of C₂H₂/(4 moles of CO₂)
moles of C₂H₂ = 1.675 moles
And finally since 1 mol occupies 22.4 L at STP we can convert back the moles of C₂H₂ into Liters.
volume of C₂H₂ = 1.675 moles * 22.4 L/(1 mol)
volume of C₂H₂ = 37.5 L
B) 2 C₂H₂ + 5 O₂ ----> 4 CO₂ + 2 H₂O
moles of CO₂ = 3.35 mol
4 moles of CO₂ = 2 moles of H₂O
1 mol of H₂O = 22.4 L
volume of H₂O = 3.35 moles of CO₂ * 2 moles of H₂O/(4 moles of CO₂) * 22.4 L/(1 mol of H₂O)
volume of H₂O = 37.5 L
C) 2 C₂H₂ + 5 O₂ ----> 4 CO₂ + 2 H₂O
volume of CO₂ = 75.0 L
4 moles of CO₂ = 5 moles of O₂
1 mol of CO₂ = 1 mol of O₂ = 22.4 L
volume of O₂ = 75.0 L of CO₂ * 1 mol of CO₂/(22.4 L of CO₂) * 5 moles of O₂/(4 moles of CO₂) * 22.4 L of O₂/(1 mol of O₂)
volume of O₂ = 93.75 L
Answer:
a) volume of C₂H₂ = 37.5 L
b) volume of H₂O = 37.5 L
c) volume of O₂ = 93.75 L