A) How to calculate the theoretical yield of water resulting from the complete combustion of 1.60 g of methane b) When the reaction was run in the lab 3.30 g of water were formed. What is the percent yield of water for this run? CH4 + 2 O2 ->CO2 + 2 H2O?

Feb 28, 2015

Once again, start with the balanced chemical equation for the combustion of methane, $C {H}_{4}$

$C {H}_{4} + 2 {O}_{2} \to C {O}_{2} + 2 {H}_{2} O$

You have a $\text{1:2}$ mole ratio between methane and water - this means that the number of moles of water produced must be twice the number of moles of methen that reacted.

The theoretical yield is the yield you'd get for a 100% reaction - i.e. when all the methane reacted and water was produced according to the aforementioned mole ratio. So, in theory, this reaction would produce

$\text{1.60 g methane" * "1 mole methane"/"16.0 g" * "2 moles water"/"1 mole methane" * "18.0 g"/"1 mole water" = "3.60 g water}$

Now, the experiment you ran produced $\text{3.30 g}$ of water; this means that not all the methane reacted $\to$ oxygen was the limiting reagent.

The reaction's percent yield for water will be

$\text{%yield" = "experimental yield"/"actual yield} \cdot 100$

$\text{%yield water" = "3.30 g"/"3.60 g" * 100 = "91.7%}$