for thr reation: 2AlCl3 + 3H2SO4= Al2(SO4)3 + 6 HCl
How many grams of H2SO4 are needed to produce 27.0 grams of Al2(SO4)3
Lillian--Here is an example stoichiometry problem I've posted. It will do about 99% of all of the simple stoichiometry problems you may have; just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
IS THE ANSWER 57.14G OF H2SO4
I don't think so. If you will post your work someone will look for the error.
23.21G
To determine the amount of H2SO4 needed to produce a given amount of Al2(SO4)3, we can use the balanced equation and stoichiometry. The coefficients in the balanced equation represent the molar ratio between reactants and products.
The balanced equation is: 2 AlCl3 + 3 H2SO4 → Al2(SO4)3 + 6 HCl
The molar ratio between H2SO4 and Al2(SO4)3 is 3:1, meaning that for every 3 moles of H2SO4, 1 mole of Al2(SO4)3 is produced.
Follow the steps below to calculate the amount of H2SO4 needed:
1. Calculate the molar mass of Al2(SO4)3:
Al: 2 * atomic mass of Al (26.98 g/mol) = 53.96 g/mol
S: 3 * atomic mass of S (32.06 g/mol) = 96.18 g/mol
O: 12 * atomic mass of O (16.00 g/mol) = 192.00 g/mol
Total molar mass of Al2(SO4)3 = 53.96 + 96.18 + 192.00 = 342.14 g/mol
2. Convert the given mass of Al2(SO4)3 to moles:
Moles of Al2(SO4)3 = given mass (g) / molar mass (g/mol)
Moles of Al2(SO4)3 = 27.0 g / 342.14 g/mol ≈ 0.079 mol
3. Use the stoichiometry from the balanced equation to find the moles of H2SO4 needed:
Moles of H2SO4 = Moles of Al2(SO4)3 * (3 moles of H2SO4 / 1 mole of Al2(SO4)3)
Moles of H2SO4 = 0.079 mol * (3/1) = 0.237 mol
4. Convert the moles of H2SO4 to grams:
Mass of H2SO4 = Moles of H2SO4 * molar mass of H2SO4
Molar mass of H2SO4 = 2 * atomic mass of H (1.01 g/mol) + atomic mass of S (32.06 g/mol) + 4 * atomic mass of O (16.00 g/mol)
= 2.02 + 32.06 + 64.00 = 98.08 g/mol
Mass of H2SO4 = 0.237 mol * 98.08 g/mol ≈ 23.29 g
Therefore, approximately 23.29 grams of H2SO4 are needed to produce 27.0 grams of Al2(SO4)3.