If a solution containing 52.044 g of mercury(II) chlorate is allowed to react completely with a solution containing 15.488 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?
I have the balanced equation right I think...
Hg(ClO3)2+Na2SO4--->HgSO4(s) + 2NaClO3
I have figured out that there are 32.34 g of solid precipitate but I don't know how to figure out the grams of the reactant in excess remaining.?
LR = limiting reagaent
ER = excess reagent.
I didn't check any of your numbers but if they are right here is what you do.
Use the coefficients to tell you just as you determined the mols of HgSO4 formed.
mols ER used = mols LR used x coefficients to convert LR to ER.
Then mols ER x molalr mass ER = g ER used.
mols ER initially - mols ER used = mols ER remaining unreacted.
To find out how many grams of solid precipitate will be formed, we first need to determine which reactant is limiting and which one is in excess. Let's calculate the number of moles for each reactant.
Step 1: Calculate the number of moles of Hg(ClO3)2:
molar mass of Hg(ClO3)2 = molar mass of Hg + 2*(molar mass of Cl + 3*(molar mass of O))
= 200.59 g/mol + 2*(35.45 g/mol + 3*(16.00 g/mol))
= 200.59 g/mol + 2*(35.45 g/mol + 48.00 g/mol)
= 200.59 g/mol + 2*(83.45 g/mol)
= 200.59 g/mol + 166.90 g/mol
= 367.49 g/mol
moles of Hg(ClO3)2 = mass / molar mass
= 52.044 g / 367.49 g/mol
= 0.1414 mol
Step 2: Calculate the number of moles of Na2SO4:
molar mass of Na2SO4 = 2*(molar mass of Na) + molar mass of S + 4*(molar mass of O)
= 2*(22.99 g/mol) + 32.07 g/mol + 4*(16.00 g/mol)
= 2*(22.99 g/mol) + 32.07 g/mol + 4*(16.00 g/mol)
= 45.98 g/mol + 32.07 g/mol + 64.00 g/mol
= 45.98 g/mol + 96.07 g/mol
= 142.05 g/mol
moles of Na2SO4 = mass / molar mass
= 15.488 g / 142.05 g/mol
= 0.109 mol
Step 3: Calculate the moles ratio between Hg(ClO3)2 and Na2SO4 using the balanced equation:
From the balanced equation: 1 mol Hg(ClO3)2 reacts with 1 mol Na2SO4
moles ratio = moles of Hg(ClO3)2 / moles of Na2SO4
= 0.1414 mol / 0.109 mol
= 1.2989
Since the moles ratio is approximately 1, it means that Hg(ClO3)2 is in excess, and Na2SO4 is the limiting reactant.
Step 4: Calculate the grams of solid precipitate formed, HgSO4:
molar mass of HgSO4 = molar mass of Hg + molar mass of S + 4*(molar mass of O)
= 200.59 g/mol + 32.07 g/mol + 4*(16.00 g/mol)
= 200.59 g/mol + 32.07 g/mol + 4*(16.00 g/mol)
= 200.59 g/mol + 32.07 g/mol + 64.00 g/mol
= 200.59 g/mol + 96.07 g/mol
= 296.66 g/mol
grams of HgSO4 = moles of HgSO4 * molar mass of HgSO4
= 0.109 mol * 296.66 g/mol
= 32.34 g
Therefore, 32.34 grams of solid precipitate (HgSO4) will be formed.
To find out the grams of the reactant in excess remaining, we need to determine the moles of the excess reactant (Hg(ClO3)2) that are not consumed.
Step 5: Calculate the remaining moles of Hg(ClO3)2:
remaining moles of Hg(ClO3)2 = moles of Hg(ClO3)2 - moles of Na2SO4 used
= 0.1414 mol - 0.109 mol
= 0.0324 mol
Step 6: Calculate the remaining grams of Hg(ClO3)2:
remaining grams of Hg(ClO3)2 = remaining moles of Hg(ClO3)2 * molar mass of Hg(ClO3)2
= 0.0324 mol * 367.49 g/mol
= 11.93 g
Therefore, there will be 11.93 grams of the reactant Hg(ClO3)2 remaining.
To determine the grams of solid precipitate formed, you need to use stoichiometry and convert the given masses of the reactants to moles. Then, compare the moles of reactants to the stoichiometric ratios in the balanced equation to find the limiting reactant.
Let's start with mercury(II) chlorate (Hg(ClO3)2):
1. Determine the moles of Hg(ClO3)2:
moles of Hg(ClO3)2 = mass of Hg(ClO3)2 / molar mass of Hg(ClO3)2
Given: mass of Hg(ClO3)2 = 52.044 g
Molar mass of Hg(ClO3)2 = molar mass of Hg + (3 x molar mass of Cl) + (6 x molar mass of O)
From the periodic table:
molar mass of Hg = 200.59 g/mol
molar mass of Cl = 35.45 g/mol
molar mass of O = 16.00 g/mol
moles of Hg(ClO3)2 = 52.044 g / (200.59 g/mol + (3 x 35.45 g/mol) + (6 x 16.00 g/mol))
Calculate the moles of sodium sulfate (Na2SO4) in a similar way using the given mass of Na2SO4 (15.488 g) and its molar mass.
Next, use the stoichiometric ratios from the balanced equation to determine which reactant is limiting:
1 mole of Hg(ClO3)2 reacts with 1 mole of HgSO4
1 mole of Na2SO4 reacts with 1 mole of HgSO4
Compare the moles of Hg(ClO3)2 and Na2SO4. If the moles of one reactant are larger than the other, that reactant is in excess, and the other reactant is the limiting reactant.
To determine the grams of the reactant in excess remaining after the reaction, you will need to calculate the moles of the excess reactant and then convert it back to mass using its molar mass.
Finally, to calculate the grams of solid precipitate formed, use the mole ratio between HgSO4 and Hg(ClO3)2 from the balanced equation and multiply it by the moles of Hg(ClO3)2 used.
I hope this explanation helps you understand how to solve the problem!