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 amount of solid precipitate formed, we need to first determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
1. Calculate the number of moles of each reactant:
- Mercury(II) acetate (Hg(CH3COO)2):
- Given mass = 33.82 g
- Molar mass of Hg(CH3COO)2 = (200.59 g/mol + 2 * (12.01 g/mol) + 4 * (1.01 g/mol) + 2 * (16.00 g/mol)) = 318.67 g/mol
- Number of moles = mass / molar mass = 33.82 g / 318.67 g/mol = 0.1061 mol
- Sodium sulfate (Na2SO4):
- Given mass = 9.718 g
- Molar mass of Na2SO4 = (2 * (22.99 g/mol) + 32.07 g/mol + 4 * (16.00 g/mol)) = 142.04 g/mol
- Number of moles = mass / molar mass = 9.718 g / 142.04 g/mol = 0.0684 mol
2. Determine the stoichiometric ratio between the reactants:
The balanced chemical equation for the reaction between mercury(II) acetate and sodium sulfate is:
Hg(CH3COO)2 + Na2SO4 -> HgSO4 + 2NaCH3COO
From the coefficients in the balanced equation, we can see that the ratio of moles between Hg(CH3COO)2 and Na2SO4 is 1:1.
3. Identify the limiting reactant:
Since the ratio of moles between the two reactants is 1:1, it is clear that the limiting reactant is the one with the lesser number of moles, which is sodium sulfate (Na2SO4) in this case (0.0684 mol).
4. Calculate the amount of solid precipitate formed:
The stoichiometric ratio between HgSO4 and Na2SO4 is also 1:1 from the balanced equation. Therefore, the number of moles of solid precipitate formed (HgSO4) is also 0.0684 mol.
To calculate the mass of the solid precipitate, we need to use its molar mass.
- Molar mass of HgSO4 = (200.59 g/mol + 32.07 g/mol + 4 * (16.00 g/mol)) = 296.65 g/mol
- Mass of solid precipitate formed = number of moles * molar mass = 0.0684 mol * 296.65 g/mol = 20.31 g
Therefore, 20.31 g of solid precipitate (mercury(II) sulfate) will be formed.
To determine the mass of the reactant in excess that remains after the reaction, we can subtract the moles of the limiting reactant (sodium sulfate) from the moles of the excess reactant (mercury(II) acetate).
- Moles of the excess reactant (Hg(CH3COO)2) = 0.1061 mol - 0.0684 mol = 0.0377 mol
To calculate the mass of the excess reactant:
- Mass of Hg(CH3COO)2 = number of moles * molar mass = 0.0377 mol * 318.67 g/mol = 12.01 g
Therefore, 12.01 g of the reactant in excess will remain after the reaction.
To solve this problem, we need to determine the limiting reactant, which is the reactant that is completely consumed in the reaction. Once we know the limiting reactant, we can use stoichiometry to calculate the amount of solid precipitate formed and the amount of the reactant in excess that remains.
Step 1: Write the balanced chemical equation for the reaction.
The balanced equation for the reaction between mercury(II) acetate (Hg(CH3COO)2) and sodium sulfate (Na2SO4) is:
Hg(CH3COO)2 + Na2SO4 -> HgSO4 + 2CH3COONa
Step 2: Calculate the number of moles of each reactant.
- Convert the mass of mercury(II) acetate to moles using its molar mass of 318.7 g/mol:
moles of Hg(CH3COO)2 = mass / molar mass = 33.82 g / 318.7 g/mol = 0.106 mol
- Convert the mass of sodium sulfate to moles using its molar mass of 142.04 g/mol:
moles of Na2SO4 = mass / molar mass = 9.718 g / 142.04 g/mol = 0.068 mol
Step 3: Determine the limiting reactant.
The limiting reactant is the reactant that produces the smallest number of moles of product. From the balanced equation, we see that one mole of Hg(CH3COO)2 reacts with one mole of Na2SO4 to produce one mole of HgSO4. Therefore, the limiting reactant is the one with the smaller number of moles.
Since the moles of mercury(II) acetate (0.106 mol) are greater than the moles of sodium sulfate (0.068 mol), the sodium sulfate is the limiting reactant.
Step 4: Calculate the mass of the solid precipitate formed.
From the balanced equation, we see that one mole of Hg(CH3COO)2 produces one mole of HgSO4. Therefore, the mass of the solid precipitate formed is the same as the molar mass of HgSO4.
The molar mass of HgSO4 is 296.6 g/mol.
mass of solid precipitate = molar mass = 296.6 g/mol
Step 5: Calculate the amount of the reactant in excess that remains.
Since sodium sulfate is the limiting reactant, it is completely consumed in the reaction. Therefore, there will be no excess of sodium sulfate remaining.
Step 6: Summarize the answers:
- The mass of solid precipitate formed is 296.6 g.
- The amount of the reactant in excess (sodium sulfate) is zero grams.
So, after the reaction, 296.6 grams of solid precipitate will be formed, and there will be no remaining sodium sulfate.