5.15 g of water
Assuming the reaction in question is the one that occurs in lead storage batteries;
Pb + PbO₂ + 2H⁺ + 2HSO₄⁻ → 2PSO₄ + 2H₂O
The reaction is balanced.
We are given 34.3 grams of PbO₂
Required to determine the mass of water that will be formed,
We can use the following steps;
To get the number of moles we divide mass by the molar mass
Molar mass of PbO₂ = 239.2 g/mol
Therefore;
Moles = 34.3 g ÷ 239.2 g/mol
= 0.143 moles
From the equation, 1 mole of PbO₂ reacts to produce 2 moles of water.
Therefore, 0.143 moles of PbO₂ would produce,
= 0.143 moles × 2
= 0.286 moles
To get mass we multiply number of moles by the molar mass
Molar mass of water = 18.02 g/mol
Therefore,
Mass = 0.286 moles × 18.02 g/mol
= 5.154 g
= 5.15 g
Therefore, 5.15 g of water will be formed
