If a solution containing 22.34 g of mercury(II) perchlorate is allowed to react completely with a solution containing 6.256 g of sodium sulfate, how many grams of solid precipitate will be formed?
To determine the amount of solid precipitate formed, we need to identify the balanced chemical equation for the reaction between mercury(II) perchlorate (Hg(ClO4)2) and sodium sulfate (Na2SO4).
The balanced chemical equation for the reaction is:
Hg(ClO4)2 + Na2SO4 → HgSO4 + 2NaClO4
From the balanced equation, we can see that 1 mole of Hg(ClO4)2 reacts with 1 mole of Na2SO4 to produce 1 mole of HgSO4.
1 mole of Hg(ClO4)2 has a molar mass of 296.409 g/mol, and 1 mole of Na2SO4 has a molar mass of 142.042 g/mol.
First, we need to calculate the number of moles of Hg(ClO4)2 and Na2SO4 in the given amounts.
Number of moles of Hg(ClO4)2 = mass of Hg(ClO4)2 / molar mass of Hg(ClO4)2
= 22.34 g / 296.409 g/mol
≈ 0.0753 mol
Number of moles of Na2SO4 = mass of Na2SO4 / molar mass of Na2SO4
= 6.256 g / 142.042 g/mol
≈ 0.0440 mol
From the balanced chemical equation, we can see that the ratio of moles of Hg(ClO4)2 to HgSO4 is 1:1. Therefore, the number of moles of HgSO4 formed will be the same as the number of moles of Hg(ClO4)2.
Number of moles of HgSO4 formed = 0.0753 mol
Finally, we can calculate the mass of the solid precipitate (HgSO4) formed using its molar mass.
Mass of HgSO4 formed = number of moles of HgSO4 formed × molar mass of HgSO4
= 0.0753 mol × (200.61 g/mol for HgSO4)
≈ 15.09 g
Therefore, approximately 15.09 grams of solid precipitate (HgSO4) will be formed in this reaction.