Final answer:
When 25.0 grams of sodium sulfate is reacted with excess lead (II) nitrate, the amount of precipitate formed can be determined using stoichiometry. The balanced chemical equation is Pb(NO3)2(aq) + Na2SO4(aq) → PbSO4(s) + 2NaNO3(aq). The mole ratio between sodium sulfate and lead (II) nitrate is 1:1, so there will be 0.1761 moles of lead (II) sulfate formed.
Step-by-step explanation:
When 25.0 grams of sodium sulfate is reacted with excess lead (II) nitrate, the amount of precipitate (product) formed can be determined using stoichiometry. We need to write the balanced chemical equation to determine the mole ratio between sodium sulfate and lead (II) nitrate. The balanced equation is:
Pb(NO3)2(aq) + Na2SO4(aq) → PbSO4(s) + 2NaNO3(aq)
The mole ratio between sodium sulfate and lead (II) nitrate is 1:1.
Given that the molar mass of sodium sulfate is 142.04 g/mol, we can convert the mass of sodium sulfate to moles:
moles of Na2SO4 = mass of Na2SO4 / molar mass of Na2SO4
moles of Na2SO4 = 25.0 g / 142.04 g/mol
moles of Na2SO4 = 0.1761 mol
Since the mole ratio between sodium sulfate and lead (II) nitrate is 1:1, the amount of moles of lead (II) nitrate consumed is also 0.1761 mol.
So, to determine the amount of precipitate (product) formed, we can use the mole ratio between lead (II) nitrate and lead (II) sulfate, which is 1:1.
Therefore, there will be 0.1761 moles of lead (II) sulfate formed.