Methane (CH4) reacts with excess oxygen (O2) to form carbon dioxide (CO2) and water (H2O). If 10.0 L of methane react at STP, calculate the volume of carbon dioxide produced.
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We can start by writing a balanced chemical equation for the reaction:
CH4 + 2O2 -> CO2 + 2H2O
According to the stoichiometry of the reaction, we can see that 1 mole of CH4 reacts to produce 1 mole of CO2.
Using the ideal gas law, we can calculate the number of moles of CH4 using the volume at STP:
PV = nRT
(1 atm) * (10.0 L) = n * (0.0821 L*atm/mol*K) * (273 K)
n = (1 atm * 10.0 L) / (0.0821 L*atm/mol*K * 273 K)
= 0.4067 mol
Therefore, 0.4067 moles of CH4 react to produce 0.4067 moles of CO2.
Now, we can use the ideal gas law again to calculate the volume of CO2 produced:
PV = nRT
V = (n * RT) / P
= (0.4067 mol * 0.0821 L*atm/mol*K * 273 K) / (1 atm)
≈ 9.05 L
Therefore, approximately 9.05 L of carbon dioxide is produced.
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