If you run the reaction to make water: 2H2 + O2 → 2H2O, starting with 3.0 moles of hydrogen gas and 2.0 moles of oxygen gas, what is the theoretical yield of your reaction (in grams) ?

The theoretical yield of the reaction is 54.06 grams

Explanation:

Step 1: Data given

Moles of hydrogen gas = 3.0 moles

Moles of oxygen gas = 2.0 mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

2H2 + O2 → 2H2O

Step 3: Calculate the limiting reactant

For 2 moles H2 consumed, we need 1 mol O2 to produce 2 moles of H2O

Hydrogen gas is the limiting reactant. It will be completely consumed. (3.0 moles).

Oxygen gas is in excess. There will react 3.0/2 = 1.5 moles of O2.

There will remain 2.0 - 1.5 = 0.5 moles

Step 4: Calculate the moles of H2O

For 2 moles of H2 we'll have 2 moles of H2O

For 3 moles H2 we'll have 3.0 moles of H2O

Step 5: Calculate theoretical yield of H2O

Mass H2O = moles H2O * molar mass H2O

Mass H2O = 3.0 moles * 18.02 g/mol

Mass H2O = 54.06 grams

The theoretical yield of the reaction is 54.06 grams

54 g is the theoretical yield

Explanation:

This is the reaction:

2H₂ + O₂ → 2H₂O

So 2 moles of hydrogen react with 1 mol of oxygen, to produce 2 mol of water.

If I have 3 moles of H₂ and 2 moles of O₂, the my limiting reactant is the hydrogen.

1 mol of O₂ react with 2 moles of H₂

S 2 mol of O₂ would react with 4 moles (I only have 3 moles)

Then, ratio is 2:2 the same as 1:1

As 2 mol of H₂ produce 2 moles of water, 3 moles of H₂ will produce 3 moles of H₂O.

This is the theoretical yield in moles. Let's convert them to mass (mol. molar mass)

3 mol. 18g/m = 54 g