4nh3 + 3o2 --> 2n2 + 6h2o
if 83.7g of o2 is mixed with 2.81 miles of nh3, what is the maximum amount of h2o that can be produced?

The balanced reaction is:

 We are given the amount of reactants to be used for the reaction. This will be the starting point of our calculation.

83.7g of O2 ( 1 mol / 32 g) = 2.62 mol O2

2.81 moles of NH3

From the balanced reaction, we have a 4:3 ratio of the reactants. The limiting reactant would be oxygen. We will use the amount for oxygen for further calculations.

2.62 mol O2 (6 mol H2O  / 3 mol O2) (18.02 g H2O / 1 mol H2O) = 94.42 g H2O

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explanation: