Answer:
1.08 g of water
Explanation:
The balanced chemical equation for the reaction of combustion of methane (CHâ‚„) is the following:
CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g)
According to the equation, 1 mol of CHâ‚„ reacts with 2 moles of Oâ‚‚. We convert from mol to grams by using the molar masses:
1 mol CHâ‚„ = (1 x 12 g/mol) + (4 x 1 g/mol) = 16 g
2 mol Oâ‚‚ = 2 x (2 x 16 g/mol) = 64 g
2 mol Hâ‚‚O = 2 x ((2 x 1 g/mol) + 16 g/mol)= 36 g
From the masses of reactants (CHâ‚„ and Oâ‚‚), we can see that the stoichiometric ratio is 64 g Oâ‚‚/16 g CHâ‚„ = 4.
First, we have to identify which reactant is the limiting reactant. We can compare the stoichiometric ratio with the actual reactants ratio (the masses of reactants we have):
1.92 g Oâ‚‚/0.80 g CHâ‚„ = 2.4
As 4>2.4, we can conclude that Oâ‚‚ is the limiting reactant.
Now, we consider the stoichiometric ratio between the limiting reactant (64 g Oâ‚‚) and the product we have to calculate (36 g Hâ‚‚O), and we multiply the ratio by the actual mass of Oâ‚‚:
1.92 g Oâ‚‚ x 36 g Hâ‚‚O/64 g Oâ‚‚ = 1.08 g
Therefore, 1.08 g of Hâ‚‚O will be produced by the chemical reaction of 0.80 g of methane with 1.92 g of oxygen.
