Determine how many moles of water vapor, H2O, will be produced at 1.00 atm and 200 C by the complete combustion of 10.5 L of methane gas, CH4.

Question

 I need help converting 200 C to the conditions of STP.

Answer 1

I don't think you need to convert toSTP for this problem. 

 

The balanced chemical equation for the combustion of methane is 

CH₄ + 2O₂ --> 2H₂O + CO₂

 

Since we have all of the conditions for the reaction, we can use the Ideal Gas Law equation to solve for the number of moles of methane:

PV = nRT

(10.5L)(1atm) = n(0.0821 L*atm/mol*K)(473K)

n = [(0.0821 L*atm/mol*K)(473K)]/[(10.5L)(1atm)]

   = 0.270 moles CH₄

 

From there, we can use the mole ratio of methane to water (2:1) in order to determine the moles of water that can theoretically be produced:

0.27 moles CH₄ x (2 moles H₂O/1 moles CH₄) = 0.540 moles H₂O