Question

If you had 17.3 g of hydrogen and 8.91 g of oxygen, which is the limiting reactant, and how many grams of water could you produce.

Answer 1

The equation for the following reaction is:

2H₂ + O₂ → 2H₂O

We have 17.3 g of H₂ gas and 8.91 g of O₂ gas. We will convert each to moles, and then convert each of those values to moles of water.

17.3 g H₂ / 2 g/mol = 8.65 moles H₂ x 2 moles H₂O/2 moles H₂ = 8.65 moles H₂O

8.91 g O₂ / 32 g/mol = 0.28 moles O₂ x 2 moles H₂O/1 mole O₂ = 0.56 moles H₂O.

The limiting reagent in this case is O₂ as it can only produce 0.56 moles of H₂O. Now we can convert the moles of H₂O to mass.

0.56 moles H₂O x 18 g/mol = 10.1 g H₂O

We can form roughly 10.1 g of H₂O.