Question

8. Calculate the mass, in grams, of CO2 that can be produced by the reaction of 75.0 g of Cha with 50.0 g of O2(g). CH_(g) + O2(g) + CO2(g) + H2O(g)

Answer 1

The balanced equation is

CH₄ + 2 O₂ -----> CO₂ + 2 H₂O

Number of moles of CH4 = 75.0 g / 16.04 g/mol = 4.68 mole

Number of moles of O2 = 50.0 g / 32.0 g/mol = 1.56 mole

From the balanced equation we can say that

1 mole of CH4 requires 2 mole of O2 so

4.68 mole of CH4 will require

= 4.68 mole of CH4 *(2 mole of O2 / 1 mole of CH4)

= 9.36 mole of O2

But we have 1.56 mole of O2 which is in short so O2 is limiting reactant

From the balanced equation we can say that

2 mole of O2 produces 1 mole of CO2 so

1.56 mole of O2 will produce

= 1.56 mole of O2 *(1 mole of CO2 / 2 mole of O2)

= 0.780 mole of CO2

mass of 1 mole of CO2 = 44.01 g so

the mass of 0.780 mole of CO2 = 34.3 g

Therefore, the mass of CO2 produced would be 34.3 g