Final answer:
To produce 119 g of water in a lead-acid battery reaction, 1368.27 g of lead would be necessary based on the stoichiometry of the balanced chemical equation.
Step-by-step explanation:
To determine the amount of lead required to make 119 g of water, we must use stoichiometry based on the balanced chemical equation provided. First, we notice that the balanced equation given in the question seems to be incorrect or incomplete. However, for the lead-acid battery reaction, the correct balanced chemical reaction is:
Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)
According to this reaction, 1 mole of lead (Pb) reacts to produce 1 mole of water (H2O). To solve this problem, we need to follow these steps:
- Calculate the moles of water produced using its molar mass (18.02 g/mol).
- Using the stoichiometry of the reaction, determine the moles of lead needed for the moles of water calculated.
- Convert the moles of lead to grams using the molar mass of lead (207 g/mol).
Using the molar mass of water (18.02 g/mol), we find that 119 g of water is 6.61 moles (119 g ÷ 18.02 g/mol = 6.61 mol). According to the stoichiometry of the reaction, 1 mole of lead produces 1 mole of water; hence, 6.61 moles of lead is required to produce 6.61 moles of water. Finally, multiplying the moles of lead by the atomic mass of lead gives us the mass of lead required:
6.61 mol × 207 g/mol = 1368.27 g
Therefore, 1368.27 g of lead would be required to produce 119 g of water in the context of this reaction.