Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation: CH4(g)+H2O(g) = CO(g) + 3H2(g)
In a particular reaction, 26.5 L of methane gas (measured at a pressure of 736 torr and a temperature of 25 C ) is mixed with 22.8 L of water vapor (measured at a pressure of 704 torr and a temperature of 125 C ). The reaction produces 27.0 L of hydrogen gas measured at STP. What is the percent yield of the reaction?

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  1. This is about three problems rolled into one. I expect it may be a limiting reagent problem so we address that first.
    1. Use PV = nRT to convert CH4 to mols. Remember temperature is in Kelvin and P is in atmospheres.
    2. Use PV = nRT to convert H2O to mols. P is in atm and T in Kelvin.
    3. Now use the coefficients in the balanced equation and
    a. convert mols CH4 to mols H2.
    b. convert mols H2O to mols H2.
    c. If a and b give the same answer, this is not a limiting reagent problem and you will use that answer throughout the remainder of the calculations. If a and b are not the same, use the smaller of the numbers and that identifies the limiting reagent.
    4. Using the appropriate value from 3a,3b, or 3c, convert mols of that compound to mols H2. That number is the theoretical yield of H2 from the reactions under these conditions.
    5. Use PV = nRT to convert 27.0 L H2 to mols. This is the actual yield.
    6. %yield = [actual yield/theoretical yield]*100 = ??
    Check my work carefully. Check my thinking.

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  2. 61.2%

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