If you burn 55.6 g of hydrogen and produce 497 g of water, how much oxygen reacted?

441.28 g Oxygen

Explanation:

The combustion of hydrogen gives water as the product. The equation for the reaction is;

2H₂ (g) + O₂ (g) → 2H₂O (l)

Mass of hydrogen = 55.6 g

Number of moles of hydrogen

Moles = Mass/Molar mass

= 55.6 g : 2.016 g/mol

= 27.8 moles

The mole ratio of Hydrogen to Oxygen is 2:1

Therefore;

Number of moles of oxygen = 27.5794 moles : 2

= 13.790 moles

Mass of oxygen gas will therefore be;

Mass = Number of moles * Molar mass

Molar mass of oxygen gas is 32 g/mol

Mass = 13.790 moles * 32 g/mol

= 441.28 gAlternatively:

Mass of hydrogen + mass of oxygen = Mass of water

Therefore;

Mass of oxygen = Mass of water - mass of hydrogen

= 497 g - 55.6 g

= 441.4 g

2H2 + O2 → 2H2O

2 mol H2 + 1mol O2 will produce 2 mol H2O

Molar mass H2O = 18.0153g/mol

497g H2O = 23.813 mol H2O produces

This will require 27.587/2 = 13.793 mol O2

Molar mass O2 = 32g/mol

13.793mol = 13.793*32 = 441.37g O2 required.

The easiest way is subtracting 55.6 from 497 which gives you 441.4.