Question

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?

Answer 1

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

Step-by-step 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