Question

Gaseous methane (CH) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water methane is mixed with 5.8 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits. (H,O). Suppose 3.85 g of

Answer 1

   CH₄ + 2O₂ ----> CO₂ + 2H₂O

Mass of methane = 3.85 g

Mass of oxygen = 5.8 g

Molar mass of methane = 16.04 g/mol

Molar mass of oxygen = 16 g/mol

No. of moles of methane = mass/molar mass = 3.85 g/16.04 g/mol = 0.24 moles

No. of moles of oxygen = mass/molar mass = 5.8 g/16 g/mol = 0.3625 moles

Molar ratio of methane and oxygen = 1 :2

1 mole of methane require 2 moles of oxygen

0.24 mole of methane require == 0.24 x 2 = 0.48 moles of oxygen but the taken moles of oxygen is 0.3625

Hence oxygen is the limiting reactant

Molar ratio of oxygen and carbon dioxide = 2:1

moles of carbon dioxide = 0.5 x 0.3625 moles = 0.18125 moles

Molar mass of carbon dioxide = 44.01 g/mol

Maximum mass of carbon dioxide = moles x molar mass = 0.18125 mol x 44.01 g/mol = 7.98 g = 8.0 g (2 significant figures)