Question

(i) Calculate the mass of CO2(g) in gram produced by the reaction between 3 mol of CH4(g) and 2 mol of

O2(g) according to the equation : CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

Answer 1

Answer: 0.1334983576 g

Explanation:

The mass must be conserved on both sides of the equation. Mass put in must must = mass put out. So we must start by finding the mass of the reactants and then the mass of 2H2O. Then we must subtractio the mass of the reactants from the mass of 2H2O which will gives us the mass of CO2.

Mass of CH4:

(12.0107)+ 4(1.00794) = 16.48246 g/mol

we are given three mols of CH4 so divide 3 mol by 16.48246 g/mol

3 mol/ 16.48246 g/mol = 0.1820116657 g CH4

Mass of 2O2:

2(15.99994) = 31.99988 g/mol

we are given 2 moles of 2O2 so divide 2 mol by 31.99988 g/mol

2 mol/ 31.99988 g/mol = 0.0625002344 g 2O2

Mass of 2H2O:

2(1.00794)+(15.99994) = 18.01582 g/mol

we are given 2 moles of 2H2O so divide 2 mol by 18.01582 g/mol

2 mol/ 18.01582 g/mol = 0.1110135425 g 2H20

Now we add up the grams on the reactatnt side and subtract that number from the mass of 2H2O:

0.1820116657 g CH4 + 0.0625002344 g 2O2 = 0.2445119001 g (total of g of reactants)

0.2445119001 g - 0.1110135425 g = 0.1334983576 g CO2