If we react 5.00 moles of Na with 5.00 miles of H2O, how many L of H2 gas will be produced?

Question

Using balanced equation: 2Na+H2O=Na2O+H2

a. If we react 5.00 moles of Na with 5.00 moles of H2O, how many L of H2 gas will be produced?

b. What is the limiting reagent?

c. How many g of excess reagent will remain?

Answer 1

First we have to find out how many moles of H₂ gas each of the reagents can produce.

It takes 2 moles of Na to make 1 mole of H₂ gas. We have 5 moles of Na, so 5 moles Na * (1 mole H₂ gas/ 2 moles Na) = 2.5 moles H₂ gas

It takes 1 moles of H₂O to make 1 mole of H₂ gas. We have 5 moles of H₂O, so 5 moles H₂O * (1 mole H₂ gas/ 1 moles H₂O) = 5 moles H₂ gas

That means Na is the limiting reagent and H₂O will be in excess by 2.5 moles (since only 2.5 moles of H₂ gas are produced and the ratio is 1:1). The molar mass of H₂O is 18.02 g/mol * 2.5 mol = 45.05 g H₂O remaining

Now for the volume of H₂ gas produced.

First we need to find the mass of H₂ gas produced. We know the molar mass of H₂ gas is 2.02 g/mol.

We made 2.5 moles, so 2.5 moles * 2.02 g/mol = 5.05 g H₂ gas.

In order to relate mass and volume, we use density. The density of H₂ gas is 0.0899 g/L.

Now we divide the mass of H₂ gas produced by the density to find the volume.

5.05 g H₂ gas / 0.0899 g/L = 56.17 L H₂ gas