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?
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To determine the grams of solid precipitate formed, we need to find the limiting reactant between mercury(II) acetate (Hg(C2H3O2)2) and sodium sulfate (Na2SO4).
1. Calculate the number of moles for each reactant:
- Moles of mercury(II) acetate = mass / molar mass
= 33.82 g / (2 * molar mass of hydrogen (H) + 2 * molar mass of carbon (C) + 6 * molar mass of oxygen (O) + 2 * molar mass of mercury (Hg))
= 33.82 g / (2 * 1.00794 g/mol + 2 * 12.011 g/mol + 6 * 15.9994 g/mol + 2 * 200.59 g/mol)
- Moles of sodium sulfate = mass / molar mass
= 9.718 g / (2 * molar mass of sodium (Na) + molar mass of sulfur (S) + 4 * molar mass of oxygen (O))
= 9.718 g / (2 * 22.9898 g/mol + 32.06 g/mol + 4 * 15.9994 g/mol)
2. Calculate the mole ratio between the two reactants:
- Divide the moles of each reactant by its stoichiometric coefficient in the balanced equation to find the mole ratio.
3. Identify the limiting reactant:
- The reactant that produces the smaller number of moles in the mole ratio is the limiting reactant. This reactant will be consumed completely in the reaction.
4. Calculate the moles and grams of solid precipitate formed:
- Use the mole ratio to find the moles of solid precipitate formed.
- Finally, multiply the moles of solid precipitate by its molar mass to get the grams.
To determine the grams of the reactant in excess remaining after the reaction, we'll use the limiting reactant calculated above.
1. Calculate the moles of the limiting reactant (the one that is consumed completely).
2. Calculate the moles of the reactant in excess using the mole ratio.
3. Multiply the moles of the excess reactant by its molar mass to find the grams remaining.
Let's go through the steps to find the answers.