Hydrogen gas and oxygen gas can be combined to produce liquid water. If 15.0 g of each gas is combined,

a. W hat mass of water will be produced, assuming the reaction runs to completion?

b. Which reactant will remain, and how many grams?

a) There will be produced 16.89 grams of H2O

b) There will remain 6.4885 moles of H2. This is 13.1 grams H2

Explanation:

Step 1: Data given

Mass of hydrogen gas = 15.0 grams

Mass of oxygen gass = 15.0 grams

Molar mass of hydrgen = 2.02 g/mol

Molar mass of oxygen = 32 g/mol

Step 2: The balanced equation:

2H2 + O2 → 2H2O

Step 3: Calculate moles of H2

Moles H2 = Mass H2 / Molar mass H2

Moles H2 = 15.00 g / 2.02 g/mol

Moles H2 = 7.426 moles

Step 4: Calculate moles of O2

Moles O2 = 15.00 g / 32g/mol

Moles O2 = 0.46875 moles

Step 5: Calculate limiting reactant

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

O2 is the limiting reactant. It will completely be consumed (0.46875 moles).

H2 is in excess. There will be consumed 0.46875 * 2 = 0.9375 moles of H2

There will remain 6.4885 moles of H2. This is 13.1 grams

Step 6: Calculate moles of H2O

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

For 0.46875 moles we will have 0.9375 moles of H2O produced.

Step 7: Calculate mass of H2O

Mass H2O = moles H2O * molar mass H2O

Mass H2O = 0.9375 moles * 18.02 g/mol

Mass H2O = 16.89 grams

There will be produced 16.89 grams of H2O