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
Now I go from grams to moles of each element?
I have figured out that 32.34 grams of solid precipitate will form but I'm not sure how to figure out the grams of the reactant in excess? I thought it was 22.01 g but that isn't right.
Yes, to determine the grams of solid precipitate formed and the grams of the reactant in excess remaining after the reaction, you would need to convert the given masses of the reactants to moles.
First, let's calculate the number of moles of mercury(II) chlorate (Hg(ClO3)2) and sodium sulfate (Na2SO4):
Molar mass of Hg(ClO3)2:
Hg: 1 * 200.59 g/mol = 200.59 g/mol
Cl: 2 * 35.45 g/mol = 70.90 g/mol
O: 6 * 16.00 g/mol = 96.00 g/mol
Total molar mass of Hg(ClO3)2: 200.59 + 70.90 + 96.00 = 367.49 g/mol
Moles of Hg(ClO3)2 = mass of Hg(ClO3)2 / molar mass of Hg(ClO3)2:
Moles of Hg(ClO3)2 = 52.044 g / 367.49 g/mol
Molar mass of Na2SO4:
Na: 2 * 22.99 g/mol = 45.98 g/mol
S: 1 * 32.07 g/mol = 32.07 g/mol
O: 4 * 16.00 g/mol = 64.00 g/mol
Total molar mass of Na2SO4: 45.98 + 32.07 + 64.00 = 142.05 g/mol
Moles of Na2SO4 = mass of Na2SO4 / molar mass of Na2SO4:
Moles of Na2SO4 = 15.488 g / 142.05 g/mol
Now that you have the moles of each reactant, you can determine the limiting reactant and calculate the moles of products formed.
Yes, you're on the right track! To determine the number of grams of solid precipitate formed and the grams of excess reactant remaining, you first need to convert the given masses of mercury(II) chlorate and sodium sulfate to moles. Here's how you can do that:
Step 1: Convert grams to moles for mercury(II) chlorate
To convert grams to moles, divide the given mass of mercury(II) chlorate (52.044 g) by its molar mass. The molar mass of mercury(II) chlorate (Hg(ClO3)2) can be calculated by summing the atomic masses of all the elements in the compound.
Molar mass of Hg(ClO3)2 = atomic mass of Hg + 2 × (atomic mass of Cl) + 6 × (atomic mass of O)
Step 2: Convert grams to moles for sodium sulfate
Similarly, divide the given mass of sodium sulfate (15.488 g) by its molar mass. The molar mass of sodium sulfate (Na2SO4) can be calculated in the same way as above.
Molar mass of Na2SO4 = 2 × (atomic mass of Na) + atomic mass of S + 4 × (atomic mass of O)
Step 3: Determine the limiting reactant
To find the limiting reactant, compare the mole amounts of mercury(II) chlorate and sodium sulfate. The reactant that produces fewer moles of the product is the limiting reactant, as it will be entirely consumed in the reaction.
Step 4: Calculate moles of solid precipitate formed
Using the balanced equation, determine the mole ratio between mercury(II) chlorate and the solid precipitate (HgSO4). For every 1 mole of HgSO4, you need 1 mole of Hg(ClO3)2.
Step 5: Calculate grams of solid precipitate formed
Multiply the moles of solid precipitate (calculated in step 4) by the molar mass of HgSO4 to get the corresponding mass in grams.
Step 6: Calculate grams of excess reactant remaining
To determine the excess reactant, subtract the moles of the limiting reactant (calculated in step 3) from the total moles of that reactant initially present. Then, multiply the remaining moles of the excess reactant by its molar mass to get the corresponding mass in grams.
Now, follow these steps to calculate the answer to the given question. Remember to use the molar masses of the respective compounds to perform the conversions.