How many grams of H2O will be produced if 16.05g of CH4 is combusted in the presence of 16.00g of O2 and the reaction yield is 90.00%?
When given grams of two reactants you should determine which, if either, is the limiting reactant as it will determine the amount of products produced.
First convert both quantities to moles:
16.05g CH4/16.1g/mol = 1.00 mol CH4
16.00g O2/32.0g/mol = 0.500 mol O2
Oxygen is the limiting reactant because the mole ratio of O2:CH4 is 2:1 according to the equation below for combustion of methane, and there are only half as many moles of O2 as CH4 present.
`CH_4 + 2O_2 -> CO_2 + 2 H_2O`
This presents another issue. When there isn't sufficient oxygen to combust a hydrocarbon it will undergo incomplete combustion, producing CO instead of CO2, according to the equation
`2CH_4 + 3 O_2 -> 2CO + 4 H_2O`
This gives us a mole ratio of 3:2 for O2:CH4, so O2 is still the limiting reactant.
Using 0.500 moles of O2 we get:
0.500 mol O2 x 4 H2O/3 O2 x 18.1g H2O/mol H2O = 12.0 g H2O
Since the yield is 90.00%, we must multiply the grams of product by 0.9000:
12 g H2O x 0.9000 = 10.8 g H2O produced