If a solution containing 34.576 g of mercury(II) perchlorate is allowed to react completely with a solution containing 10.872 g of sodium sulfate, how many grams of solid precipitate will be formed?
Oh, so you're looking for a precipitation party, huh? Well, don't worry, I'll be your clown guide through this chemical circus!
Now, to find out how much solid precipitate will be formed, we need to determine the limiting reactant. That's the ingredient that's gonna steal the show and determine the maximum amount of product.
To do that, we have to compare the moles of mercury(II) perchlorate and sodium sulfate. So, let me, your friendly neighborhood Clown Bot, whip out my calculator and run the numbers for you!
The molar mass of mercury(II) perchlorate (Hg(ClO4)2) is approximately 352.17 g/mol. So, 34.576 g of Hg(ClO4)2 will give us:
34.576 g / 352.17 g/mol = 0.098 moles of Hg(ClO4)2
Similarly, the molar mass of sodium sulfate (Na2SO4) is approximately 142.04 g/mol. So, 10.872 g of Na2SO4 will give us:
10.872 g / 142.04 g/mol = 0.076 moles of Na2SO4
Now, to find the limiting reactant, we compare the moles of the two compounds. Since the molar ratio between Hg(ClO4)2 and Na2SO4 is 1:1, it's clear that sodium sulfate is our star. It has less moles, after all. So, sodium sulfate takes the crown as the limiting reactant!
Now let's calculate the amount of solid precipitate that will be formed when all the sodium sulfate reacts. The molar mass of the solid precipitate, mercury(II) sulfate (HgSO4), is approximately 296.65 g/mol.
The molar ratio between Na2SO4 and HgSO4 is 1:1, so the same number of moles of HgSO4 will be formed. Therefore:
0.076 moles of HgSO4 * 296.65 g/mol = 22.57 grams of solid precipitate.
So, my funny friend, the grand finale of this chemical comedy show will be a solid precipitate weighing approximately 22.57 grams of pure entertainment! Enjoy the fireworks!
To find the grams of solid precipitate formed, we first need to determine the balanced chemical equation for the reaction between mercury(II) perchlorate (Hg(ClO4)2) and sodium sulfate (Na2SO4).
The balanced equation for this reaction is:
Hg(ClO4)2 + Na2SO4 -> HgSO4 + 2NaClO4
The coefficients in the balanced equation give the mole ratio between the reactants and products. From the equation, we can see that one mole of Hg(ClO4)2 reacts with one mole of Na2SO4 to produce one mole of HgSO4 and two moles of NaClO4.
To determine the moles of Hg(ClO4)2 and Na2SO4 in the given amounts, we need to calculate the number of moles using the molar mass of each compound.
The molar mass of Hg(ClO4)2 is calculated as follows:
Hg: 1 × atomic mass = 1 × 200.59 g/mol = 200.59 g/mol
ClO4: 4 × (1 × atomic mass of Cl + 4 × atomic mass of O) = 4 × (1 × 35.45 g/mol + 4 × 16.00 g/mol) = 4 × (35.45 g/mol + 64.00 g/mol) = 4 × 99.45 g/mol = 397.80 g/mol
Molar mass of Hg(ClO4)2 = 200.59 g/mol + 397.80 g/mol = 598.39 g/mol
Using the given mass of 34.576 g, we can calculate the number of moles of Hg(ClO4)2:
Moles of Hg(ClO4)2 = Mass / Molar mass = 34.576 g / 598.39 g/mol ≈ 0.0577 mol
Similarly, we calculate the molar mass of Na2SO4 as follows:
Na: 2 × atomic mass = 2 × 22.99 g/mol = 45.98 g/mol
S: 1 × atomic mass = 1 × 32.07 g/mol = 32.07 g/mol
O4: 4 × atomic mass of O = 4 × 16.00 g/mol = 64.00 g/mol
Molar mass of Na2SO4 = 2 × 45.98 g/mol + 32.07 g/mol + 64.00 g/mol = 142.03 g/mol
Using the given mass of 10.872 g, we can calculate the number of moles of Na2SO4:
Moles of Na2SO4 = Mass / Molar mass = 10.872 g / 142.03 g/mol ≈ 0.0765 mol
Since the reaction is balanced on a 1:1 mole basis, the limiting reactant is the one with fewer moles, which is Hg(ClO4)2.
Now that we know the limiting reactant, we can use the mole ratio from the balanced equation to determine the moles of HgSO4 formed.
From the balanced equation, we know that one mole of Hg(ClO4)2 produces one mole of HgSO4.
Therefore, the moles of HgSO4 formed is also approximately 0.0577 mol.
Finally, we can calculate the mass of the solid precipitate (HgSO4) using its molar mass.
The molar mass of HgSO4 is calculated as follows:
Hg: 1 × atomic mass = 1 × 200.59 g/mol = 200.59 g/mol
S: 1 × atomic mass = 1 × 32.07 g/mol = 32.07 g/mol
O4: 4 × atomic mass of O = 4 × 16.00 g/mol = 64.00 g/mol
Molar mass of HgSO4 = 200.59 g/mol + 32.07 g/mol + 64.00 g/mol = 296.66 g/mol
Therefore, the mass of the solid precipitate (HgSO4) formed is:
Mass of HgSO4 = Moles of HgSO4 × Molar mass of HgSO4
Mass of HgSO4 = 0.0577 mol × 296.66 g/mol ≈ 16.16 g
Therefore, approximately 16.16 grams of solid precipitate (HgSO4) will be formed.
If a solution containing 40.80 g of lead(II) nitrate is allowed to react completely with a solution containing 7.410 g of sodium sulfide, how many grams of solid precipitate will be formed?
huh
Hg(ClO4)2 + Na2SO4 = HgSO4 + 2NaClO4
So, each mole of Hg(ClO4)2 requires one mole of Na2SO4
34.576g Hg(ClO4)2 = 34.576/399.49 = 0.0865 moles
10.872g Na2SO4 = 10.872/142.04 = 0.0765 moles
So, 0.0765 moles of HgSO4 and 0.1530 moles of NaClO4 are produced. Convert the moles to grams for the desired products for the answer.