Al2(SO4)3+6KOH-->2Al(OH)3+3K2SO4
a)How many grams of potassium hydroxide are needed to completely react with 870 mg of aluminum sulfate?
b)If you want to make 7.00 grams of potassium sulfate, how many grams of aluminum sulfate would you need to react?
c)How many grams of aluminum hydroxide would you make if you completely react 1.23*10^-5 moles of aluminum sulfate?
convert grams to moles
Each mole of Al2(SO4)3 requires 6 moles of KOH and produces 2 moles of Al(OH)3
Now just plug in your numbers and chug out the answers.
To answer these questions, we need to use the balanced chemical equation for the reaction:
Al2(SO4)3 + 6KOH → 2Al(OH)3 + 3K2SO4
a) To find how many grams of potassium hydroxide (KOH) are needed to completely react with 870 mg of aluminum sulfate (Al2(SO4)3), we first need to calculate the number of moles of aluminum sulfate.
Given:
Mass of aluminum sulfate (Al2(SO4)3) = 870 mg
Molar mass of aluminum sulfate (Al2(SO4)3):
Al: 2(26.98) = 53.96 g/mol
S: 3(32.07) = 96.21 g/mol
O: 12(16.00) = 192.00 g/mol
Total molar mass = 53.96 + 96.21 + 192.00 = 342.17 g/mol
Moles of aluminum sulfate (Al2(SO4)3) = mass / molar mass
Moles of aluminum sulfate = 870 mg / 342.17 g/mol
Moles of aluminum sulfate ≈ 0.00254 mol
From the balanced chemical equation, we need 6 moles of KOH to react with 1 mole of aluminum sulfate. Therefore, we can calculate the grams of KOH needed:
Molar mass of potassium hydroxide (KOH):
K: 39.10 g/mol
O: 16.00 g/mol
H: 1(1.01) = 1.01 g/mol
Total molar mass = 39.10 + 16.00 + 1.01 = 56.11 g/mol
Grams of KOH needed = moles of aluminum sulfate * (6 moles of KOH / 1 mole of aluminum sulfate) * molar mass of KOH
Grams of KOH needed = 0.00254 mol * (6 mol / 1 mol) * 56.11 g/mol
b) To find how many grams of aluminum sulfate (Al2(SO4)3) are needed to react and produce 7.00 grams of potassium sulfate (K2SO4), we will follow a similar approach.
Given:
Mass of potassium sulfate (K2SO4) = 7.00 g
Molar mass of potassium sulfate (K2SO4):
K: 2(39.10) = 78.20 g/mol
S: 32.07 g/mol
O: 4(16.00) = 64.00 g/mol
Total molar mass = 78.20 + 32.07 + 64.00 = 174.27 g/mol
Moles of potassium sulfate (K2SO4) = mass / molar mass
Moles of potassium sulfate = 7.00 g / 174.27 g/mol
Moles of potassium sulfate ≈ 0.04017 mol
From the balanced chemical equation, we need 3 moles of K2SO4 to react with 1 mole of aluminum sulfate. Therefore, we can calculate the grams of aluminum sulfate needed:
Grams of aluminum sulfate needed = moles of potassium sulfate * (1 mole of aluminum sulfate / 3 moles of potassium sulfate) * molar mass of aluminum sulfate
c) To find how many grams of aluminum hydroxide (Al(OH)3) would be produced when 1.23 × 10^-5 moles of aluminum sulfate (Al2(SO4)3) is completely reacted, we use the balanced chemical equation.
From the balanced chemical equation, 1 mole of aluminum sulfate produces 2 moles of aluminum hydroxide.
Molar mass of aluminum hydroxide:
Al: 26.98 g/mol
O: 3(16.00) = 48.00 g/mol
H: 3(1.01) = 3.03 g/mol
Total molar mass = 26.98 + 48.00 + 3.03 = 77.01 g/mol
Grams of aluminum hydroxide produced = moles of aluminum sulfate * (2 moles of aluminum hydroxide / 1 mole of aluminum sulfate) * molar mass of aluminum hydroxide
Now, you can substitute the given values and calculate the results for each question.
To answer these questions, we need to balance the chemical equation first:
2Al2(SO4)3 + 6KOH -> 4Al(OH)3 + 3K2SO4
Now, let's proceed to solve each question step by step:
a) How many grams of potassium hydroxide are needed to completely react with 870 mg of aluminum sulfate?
To solve this, we need to use the molar ratios between aluminum sulfate (Al2(SO4)3) and potassium hydroxide (KOH) from the balanced equation.
1 mole of Al2(SO4)3 reacts with 6 moles of KOH.
First, we convert 870 mg of aluminum sulfate to grams:
870 mg = 0.870 g
Next, we convert grams of Al2(SO4)3 to moles by using its molar mass:
Molar mass of Al2(SO4)3 = 2(26.98 g/mol) + 3(32.07 g/mol) + 12(16.00 g/mol) = 342.15 g/mol
Moles of Al2(SO4)3 = mass (g) / molar mass (g/mol) = 0.870 g / 342.15 g/mol = 0.00254 mol
Now, using the mole ratio, we can calculate the moles of KOH:
Moles of KOH = 6 * moles of Al2(SO4)3 = 6 * 0.00254 mol = 0.01524 mol
Finally, convert moles of KOH to grams by using its molar mass:
Molar mass of KOH = 39.10 g/mol + 16.00 g/mol + 1.01 g/mol = 56.11 g/mol
Grams of KOH = moles of KOH * molar mass of KOH = 0.01524 mol * 56.11 g/mol = 0.856 g
Therefore, approximately 0.856 grams of potassium hydroxide are needed to completely react with 870 mg of aluminum sulfate.
b) If you want to make 7.00 grams of potassium sulfate, how many grams of aluminum sulfate would you need to react?
Similarly, we will use the balanced equation and the molar ratios to solve this question.
Looking at the balanced equation:
3 moles of K2SO4 react with 2 moles of Al2(SO4)3
First, convert grams of K2SO4 to moles by using its molar mass:
Molar mass of K2SO4 = 39.10 g/mol + 32.07 g/mol + 4(16.00 g/mol) = 174.25 g/mol
Moles of K2SO4 = mass (g) / molar mass (g/mol) = 7.00 g / 174.25 g/mol = 0.04015 mol
Now, using the mole ratio, we can calculate the moles of Al2(SO4)3:
Moles of Al2(SO4)3 = (2/3) * moles of K2SO4 = (2/3) * 0.04015 mol = 0.02676 mol
Finally, convert moles of Al2(SO4)3 to grams by using its molar mass:
Grams of Al2(SO4)3 = moles of Al2(SO4)3 * molar mass of Al2(SO4)3 = 0.02676 mol * 342.15 g/mol = 9.15 g
Therefore, approximately 9.15 grams of aluminum sulfate would be needed to react with 7.00 grams of potassium sulfate.
c) How many grams of aluminum hydroxide would you make if you completely react 1.23 x 10^-5 moles of aluminum sulfate?
Again, we will use the balanced equation and the molar ratios to find the solution.
From the balanced equation, we know that:
2 moles of Al2(SO4)3 react to produce 4 moles of Al(OH)3
First, calculate the moles of Al(OH)3 produced from the given moles of Al2(SO4)3:
Moles of Al(OH)3 = (4/2) * moles of Al2(SO4)3 = (4/2) * 1.23 x 10^-5 mol = 2.46 x 10^-5 mol
Next, calculate the mass of Al(OH)3 using its molar mass:
Molar mass of Al(OH)3 = 2(26.98 g/mol) + 3(1.01 g/mol) + 16.00 g/mol = 78.00 g/mol
Mass of Al(OH)3 = moles of Al(OH)3 * molar mass of Al(OH)3 = 2.46 x 10^-5 mol * 78.00 g/mol = 1.92 x 10^-3 g
Therefore, approximately 1.92 x 10^-3 grams of aluminum hydroxide would be produced if you completely react 1.23 x 10^-5 moles of aluminum sulfate.