12.4g of methane mixed with 30g of oxygen maximum mass of water?

Question

12.4g of methane mixed with 30g of oxygen maximum mass of water?

Answer 1

Methane = CH₄

CH₄ + 2O₂ ==> CO₂ + 2H₂O ... balanced equation

Find which reactant is limiting:

12.4 g CH₄ x 1 mole CH4/16 g = 0.775 moles CH₄ present

30 g O₂ x 1 mole O₂/32 g = 0.9375 moles O₂ present

BUT since it takes TWO moles of O₂ for each 1 mole CH₄, there isn't enough O₂, so it is limiting.

Maximum mass of H₂O produced will be dependent on O₂. Thus...

0.9375 moles O₂ x 2 moles H₂O / 2 moles O₂ x 18 g H₂O/mole H₂O = 16.9 g = 17 g H₂O (2 s.f. assuming th 30 g of O2 was actually 30. g )