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

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%”##