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Pena Brooks
04/26/2024 · Primary School

If a solution containing 53.75 g of mercury(II) nitrate is allowed to react completely with a solution containing 14.334 g of sodium sulfate according to the equation below. \[ \mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \longrightarrow 2 \mathrm{NaNO}_{3}(\mathrm{aq})+\mathrm{HgSO}_{4}(\mathrm{~s}) \] How many grams of solid precipitate will be formed? mass: How many grams of the reactant in excess will remain after the reaction? mass:

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To determine the mass of the solid precipitate formed and the mass of the reactant in excess, we need to follow these steps: 1. **Calculate the molar masses of the reactants:** - Molar mass of \( \text{Hg(NO}_3\text{)}_2 \): - Hg: 200.59 g/mol - N: 14.01 g/mol (2 N in the formula) - O: 16.00 g/mol (6 O in the formula) - Molar mass = \( 200.59 + 2(14.01) + 6(16.00) = 200.59 + 28.02 + 96.00 = 324.61 \, \text{g/mol} \) - Molar mass of \( \text{Na}_2\text{SO}_4 \): - Na: 22.99 g/mol (2 Na in the formula) - S: 32.07 g/mol - O: 16.00 g/mol (4 O in the formula) - Molar mass = \( 2(22.99) + 32.07 + 4(16.00) = 45.98 + 32.07 + 64.00 = 142.05 \, \text{g/mol} \) 2. **Convert grams of reactants to moles:** - Moles of \( \text{Hg(NO}_3\text{)}_2 \): \[ \text{Moles of } \text{Hg(NO}_3\text{)}_2 = \frac{53.75 \, \text{g}}{324.61 \, \text{g/mol}} \approx 0.165 \, \text{mol} \] - Moles of \( \text{Na}_2\text{SO}_4 \): \[ \text{Moles of } \text{Na}_2\text{SO}_4 = \frac{14.334 \, \text{g}}{142.05 \, \text{g/mol}} \approx 0.101 \, \text{mol} \] 3. **Determine the limiting reactant:** The balanced equation shows that 1 mole of \( \text{Hg(NO}_3\text{)}_2 \) reacts with 1 mole of \( \text{Na}_2\text{SO}_4 \). - From the moles calculated: - \( \text{Hg(NO}_3\text{)}_2 \): 0.165 mol - \( \text{Na}_2\text{SO}_4 \): 0.101 mol Since \( \text{Na}_2\text{SO}_4 \) has fewer moles, it is the limiting reactant. 4. **Calculate the mass of the precipitate \( \text{HgSO}_4 \) formed:** According to the balanced equation, 1 mole of \( \text{Na}_2\text{SO}_4 \) produces 1 mole of \( \text{HgSO}_4 \). Therefore, 0.101 moles of \( \text{Na}_2\text{SO}_4 \) will produce 0.101 moles of \( \text{HgSO}_4 \). - Molar mass of \( \text{HgSO}_4 \): - Hg: 200.59 g/mol - S: 32.07 g/mol - O: 16.00 g/mol (4 O in the formula) - Molar mass = \( 200.59 + 32.07 + 4(16.00) = 200.59 + 32.07 + 64.00 = 296.66 \, \text{g/mol} \) - Mass of \( \text{HgSO}_4 \): \[ \text{Mass of } \text{HgSO}_4 = 0.101 \, \text{mol} \times 296.66 \, \text{g/mol} \approx 30.00 \, \text{g} \] 5. **Calculate the mass of the reactant in excess:** - Moles of \( \text{Hg(NO}_3\text{)}_2 \) used: Since 0.101 moles of \( \text{Na}_2\text{SO}_4 \) are used, the same amount of \( \text{Hg(NO}_3\text{)}_2 \) is used. - Moles of \( \text{

Quick Answer

The mass of the solid precipitate formed is approximately 30.00 g. The mass of the reactant in excess is approximately 20.00 g.
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