If a solution containing 88.445 g of mercury(II) perchlorate is allowed to react completely with a solution containing 13.180 g of sodium sulfide, how many grams of solid precipitate will be formed?
How many grams of the reactant in excess will remain after the reaction?
To find the grams of solid precipitate formed, we need to determine the limiting reactant first.
Step 1: Write out the balanced chemical equation for the reaction:
Hg(ClO4)2 + Na2S → HgS + 2NaClO4
Step 2: Calculate the number of moles for each reactant:
- Moles of Hg(ClO4)2 = mass/molar mass = 88.445 g / (200.59 g/mol) = 0.4406 mol
- Moles of Na2S = mass/molar mass = 13.180 g / (78.045 g/mol) = 0.1686 mol
Step 3: Use the mole ratio from the balanced equation to determine the limiting reactant:
The mole ratio is 1:1 between Hg(ClO4)2 and Na2S.
Since the ratio of moles of Hg(ClO4)2 (0.4406 mol) to moles of Na2S (0.1686 mol) is greater than 1, Hg(ClO4)2 is in excess, and Na2S is the limiting reactant.
Step 4: Calculate the moles of solid precipitate formed:
From the balanced equation, we see that the mole ratio between Na2S and HgS is 1:1.
Therefore, the moles of HgS formed is also 0.1686 mol.
Step 5: Calculate the mass of solid precipitate formed:
Mass = moles x molar mass = 0.1686 mol x (232.66 g/mol) = 39.18 g
So, 39.18 grams of solid precipitate (HgS) will be formed.
To calculate the grams of the reactant in excess that will remain after the reaction, we need to find the moles of excess reactant.
Step 6: Calculate the moles of excess Hg(ClO4)2:
Moles of Na2S used = moles of HgS formed = 0.1686 mol
Moles of Hg(ClO4)2 remaining = initial moles - moles used = 0.4406 mol - 0.1686 mol = 0.272 mol
Step 7: Calculate the mass of excess Hg(ClO4)2 remaining:
Mass = moles x molar mass = 0.272 mol x (200.59 g/mol) = 54.51 g
So, 54.51 grams of the reactant Hg(ClO4)2 will remain.
To determine the grams of solid precipitate formed, we need to find the limiting reactant first. The limiting reactant is the reactant that will be completely consumed and determines the amount of product formed.
Let's start by calculating the molar masses of mercury(II) perchlorate (Hg(ClO4)2) and sodium sulfide (Na2S).
The molar mass of Hg(ClO4)2 = atomic mass of Hg + 2 * (atomic mass of Cl + 4 * atomic mass of O)
= 200.59 g/mol + 2 * (35.45 g/mol + 4 * 16.00 g/mol)
= 200.59 g/mol + 2 * (35.45 g/mol + 64.00 g/mol)
= 200.59 g/mol + 2 * (99.45 g/mol)
= 200.59 g/mol + 198.90 g/mol
= 399.49 g/mol
The molar mass of Na2S = 2 * (atomic mass of Na) + atomic mass of S
= 2 * 22.99 g/mol + 32.07 g/mol
= 45.98 g/mol + 32.07 g/mol
= 78.05 g/mol
Next, we need to calculate the number of moles of each reactant.
Number of moles of Hg(ClO4)2 = mass of Hg(ClO4)2 / molar mass of Hg(ClO4)2
= 88.445 g / 399.49 g/mol
≈ 0.2213 mol
Number of moles of Na2S = mass of Na2S / molar mass of Na2S
= 13.180 g / 78.05 g/mol
= 0.1688 mol
Now, we can compare the mole ratios of the reactants to determine the limiting reactant.
From the balanced chemical equation:
Hg(ClO4)2 + Na2S -> HgS + 2 NaClO4
According to the equation, the stoichiometric ratio is 1:1 for Hg(ClO4)2 and Na2S. This means that for every 1 mol of Hg(ClO4)2, we need 1 mol of Na2S to react completely.
Since we have 0.2213 mol Hg(ClO4)2 and 0.1688 mol Na2S, it is clear that Na2S is the limiting reactant because we don't have enough of it to react with the available Hg(ClO4)2 completely.
To find the grams of solid precipitate formed (HgS), we need to use the mole ratio between HgS and Na2S.
From the balanced chemical equation, the molar mass of HgS = atomic mass of Hg + atomic mass of S
= 200.59 g/mol + 32.07 g/mol
= 232.66 g/mol
Number of moles of HgS = number of moles of Na2S (since the stoichiometric ratio is 1:1)
= 0.1688 mol
Mass of HgS = number of moles of HgS * molar mass of HgS
= 0.1688 mol * 232.66 g/mol
= 39.16 g
Therefore, 39.16 grams of solid precipitate (HgS) will be formed.
Now, let's calculate the mass of the reactant in excess that remains after the reaction.
Number of moles of excess Hg(ClO4)2 = number of moles of Hg(ClO4)2 - number of moles of Na2S
= 0.2213 mol - 0.1688 mol
= 0.0525 mol
Mass of excess Hg(ClO4)2 = number of moles of excess Hg(ClO4)2 * molar mass of Hg(ClO4)2
= 0.0525 mol * 399.49 g/mol
= 20.97 g
Therefore, after the reaction, 20.97 grams of the reactant Hg(ClO4)2 will remain in excess.