PLEASE HELP
how many grams of AL2(SO4)3 can be made by reacting 50ml 2.8M H2SO4 with 7.4g of AL(OH)3?
3 H2SO4 + 2 Al (OH)3 ----> 6 H20 + AL2(SO4)3
This is a limiting reagent (LR) problem. You know that afbecasue amounts are given for BOTH reactants. Here are the steps. Print this out; it will work all of your LR problems.
1. Write and balance the equation. You've done that.
2. Convert grams to mols
mols = grams/molar mass. Do that for Al(OH)3 AND for H2SO4
3. Using the coefficients in the balanced equation, convert mols Al(OH)3 to mols Al2(SO4)3
4. Do the same and convert mols H2SO4 to mols Al2(SO4)3.
5. It is likely that the mols Al2(SO4)3 will NOT be the same. The correct value in LR problems is ALWAYS the smaller of the two and the reagent responsible for that smaller number is the LR.
6. Now convert that value into grams. g = mols x molar mass = ?
Post your work if you get stuck
Thank you so much for your help!
One question,
For the 50ml 2.8M H2SO4 conversion, how do i convert it? Do i go for 2.8M to molar mass or 50ml to molar mass? and what do i do with the other unit?
and do i do step 2 and 3 in one line? like the molar mass conversion and the coefficient conversion?
You can do it all in one line but I think it's easier to do it in two in order not to be confused. To convert 50 mL of 2.8 M H2SO4 to mols (sorry I forgot that ) you do
0.05 x 2.8 = ?mols H2SO4.
To find out how many grams of AL2(SO4)3 can be made in this reaction, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be obtained.
First, let's calculate the number of moles of each reactant:
1. Convert the volume of H2SO4 from milliliters (ml) to liters (L):
50 ml H2SO4 * (1 L / 1000 ml) = 0.05 L H2SO4
2. Use the molarity (M) and volume (L) to calculate the number of moles of H2SO4:
Moles of H2SO4 = Molarity * Volume
Moles of H2SO4 = 2.8 M * 0.05 L = 0.14 moles H2SO4
3. Use the molar mass of AL(OH)3 to calculate the number of moles of AL(OH)3:
The molar mass of AL(OH)3 = (27 g/mol + 3 * 16 g/mol + 3 * 1 g/mol) = 78 g/mol
Moles of AL(OH)3 = Mass / Molar mass
Moles of AL(OH)3 = 7.4 g / 78 g/mol = 0.094 moles AL(OH)3
By examining the balanced chemical equation, we can see that the stoichiometric ratio between H2SO4 and AL(OH)3 is 3:2. This means that 3 moles of H2SO4 react with 2 moles of AL(OH)3 to produce 1 mole of AL2(SO4)3.
4. Calculate the number of moles of AL2(SO4)3 that can be formed from the given moles of reactants:
By the stoichiometric ratio, the moles of H2SO4 and AL(OH)3 should be in a 3:2 ratio.
Since the ratio is 3:2, to form AL2(SO4)3, we need to check which reactant is limiting.
Moles of AL2(SO4)3 from H2SO4 = (0.14 moles H2SO4) * (1 mole AL2(SO4)3 / 3 moles H2SO4) = 0.047 moles AL2(SO4)3
Moles of AL2(SO4)3 from AL(OH)3 = (0.094 moles AL(OH)3) * (1 mole AL2(SO4)3 / 2 moles AL(OH)3) = 0.047 moles AL2(SO4)3
From the calculations, we can see that both H2SO4 and AL(OH)3 produce the same number of moles of AL2(SO4)3, which means that neither reactant is in excess. Therefore, both reactants are the limiting reactants.
Now, to find the mass of AL2(SO4)3 produced:
5. Calculate the molar mass of AL2(SO4)3:
AL2(SO4)3 has a molar mass of (2 * 27 g/mol + 3 * 32 g/mol + 12 * 16 g/mol) = 342 g/mol
6. Calculate the mass of AL2(SO4)3 produced from the limiting reactant:
Mass of AL2(SO4)3 = Moles of AL2(SO4)3 * Molar mass
Mass of AL2(SO4)3 = 0.047 moles * 342 g/mol ≈ 16.074 g
Therefore, approximately 16.074 grams of AL2(SO4)3 can be produced by reacting 50 mL of 2.8 M H2SO4 with 7.4 g of AL(OH)3.