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QUESTION

# 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 -&gt;CO2 + 2 H2...

Once again, start with the balanced chemical equation for the combustion of methane, CH_4

CH_4 + 2O_2 -> CO_2 + 2H_2O

You have a "1:2" 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

"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 "3.30 g" of water; this means that not all the methane reacted -> oxygen was the .

The reaction's for water will be

"%yield" = "experimental yield"/"actual yield" * 100

"%yield water" = "3.30 g"/"3.60 g" * 100 = "91.7%"