Consider this reaction: 2H2(g) + O2(g) = 2H20(g) First if 5.74g of H2 reacted with 32.0 of O2, determine the limiting reactant. Second, what theoretical mass of product of H20 would be produced

Question

Step by step where each number came from and why you are getting those numbers.

Specifically answer what is the limiting reactant and state the mass of product for H2O

Answer 1

2H₂(g) + O₂(g) ==> 2H₂O(g) ... Balanced equation

1.) Find moles of each reactant:

moles H₂ = 5.74 g H₂ x 1 mole H₂/2.00 g = 2.87 moles H₂

moles O₂ = 32.0 g O₂ x 1 mole O₂/32.0 g = 1.00 moles O₂

2.) Find limiting reactant:

There are a couple of ways to do this. The easiest is to divide moles of each reactant found in step 1, by the coefficient in the balanced equation and see which is lowest. Thus,....

For H₂, you have 2.87 / 1 = 1.44

For O₂, you have 1.00 / 1 = 1.00 which is less than 1.44 so O₂ is limiting.

The other way to find the limiting reactant is to calculate the theoretical yield from both reactants and see which is less. You do this using the mole ratios (coefficients) in the balanced equation. Thus,....

For H₂, you have 2.87 mol H₂ x 2 mol H₂O/2 mol H₂ = 2.87 moles H₂O theoretical yield

For O_2, you have 1.00 mol O₂ x 2 mol H₂O/1 mol O₂ = 2.00 moles H₂O theoretical yield

This shows that O₂ is limiting.

3.) Find theoretical mass of H₂O:

Using the limiting reactant, calculate mass of H₂O using the mole ratios of the balanced equation. Thus,...

1.00 moles O₂ x 2 moles H₂O/1 mole O₂ x 18 g H₂O/mole H₂O = 36.0 g H₂O theoretical yield