Question

Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O) Suppose 8.5 g of methane is mixed with 64.4 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2 sig figs.

Answer 1

The balanced equation is

CH₄ + 2 O₂ -----> CO₂ + 2 H₂O

Number of moles of CH4 = 8.5 g / 16.04 g/mol = 0.530 mol

Number of moles of O2 = 64.4 g / 32.0 g/mol = 2.01 mol

From the balanced equation we can say that

1 mole of CH4 requires 2 mole of O2 so

0.530 mole of CH4 will require

= 0.530 mole of CH4 *( 2 mole of O2 / 1 mole of CH4)

= 1.06 mole of O2

But we have 2.01 mole of O2 which is in excess so O2 is the excess reactant

and CH4 is the limiting reactant

From the balanced equation we can say that

1 mole of CH4 produces 2 mole of H2O so

0.530 mole of CH4 will produce

= 0.530 mole of CH4 *(2 mole of H2O / 1 mole of CH4)

= 1.06 mole of H2O

mass of 1 mole of H2O = 18.015 g

so the mass of the 1.06 mole of H2O = 19 g

Therefore, the mass of H2O produced would be 19 g