If a solution containing 33.82 g of mercury(II) acetate is allowed to react completely with a solution containing 9.718 g of sodium sulfate, how many grams of solid precipitate will be formed?
How many grams of the reactant in excess will remain after the reaction?
To determine the number of grams of solid precipitate formed, we need to first write a balanced chemical equation for the reaction between mercury(II) acetate and sodium sulfate. The equation is as follows:
Hg(CH3COO)2 + Na2SO4 → HgSO4 + 2NaCH3COO
From the balanced equation, we can see that one mole of mercury(II) acetate reacts with one mole of sodium sulfate to produce one mole of solid precipitate, mercury(II) sulfate. The molar mass of mercury(II) sulfate (HgSO4) is 296.67 g/mol.
Now, let's calculate the number of moles of mercury(II) acetate and sodium sulfate:
Moles of mercury(II) acetate = mass / molar mass
Moles of mercury(II) acetate = 33.82 g / 296.67 g/mol = 0.114 moles
Moles of sodium sulfate = mass / molar mass
Moles of sodium sulfate = 9.718 g / 142.04 g/mol = 0.068 moles
Since the stoichiometry of the reaction is 1:1, and there is no limiting reactant in this case, both reactants will be completely consumed. Therefore, the number of moles of solid precipitate formed will be equal to the number of moles of mercury(II) acetate used, which is 0.114 moles.
To calculate the mass of solid precipitate formed, we can use the following equation:
Mass of solid precipitate = moles of solid precipitate × molar mass of solid precipitate
Mass of solid precipitate = 0.114 moles × 296.67 g/mol = 33.79 g
Therefore, approximately 33.79 grams of solid precipitate will be formed.
To find out how many grams of the reactant in excess will remain after the reaction, we need to compare the number of moles of the two reactants. In this case, sodium sulfate is the reactant in excess since its moles (0.068 moles) are less than the moles of mercury(II) acetate (0.114 moles).
To calculate the mass of sodium sulfate leftovers, we can use the following equation:
Mass of sodium sulfate in excess = moles of sodium sulfate in excess × molar mass of sodium sulfate
Mass of sodium sulfate in excess = (0.114 moles - 0.068 moles) × 142.04 g/mol = 6.433 g
Therefore, approximately 6.433 grams of sodium sulfate will remain after the reaction.