In a solution of  Al2(SO4)3, (aq), the Al3+ concentration is 0.12M. What is the mass of Al2(SO4)3 is in 50.0 mL of this solution?

a) How many moles of Al₂(SO₄)₃ are required to make 45 mL of a 0.080 M Al₂(SO₄)₃ solution?
(Hint, molarity is moles per liter. You know the molarity. You are given the number of mL's, which can be converted to liters. Just set it up so the units cancel and you get "moles".)
_____________ moles Al₂(SO₄)₃

b) What is the molar mass of Al₂(SO₄)₃, to the nearest gram?
_____________ [grams/mole]

c) What mass of Al₂(SO₄)₃ is required to make 45 mL of a 0.080 M Al₂(SO₄)₃ solution?
(If you've got moles and molecular weight you can get to grams.)
______________ g Al₂(SO₄)₃

d) How many moles of aluminum ions are present in the solution?
(This is easy. You know how many moles of Al₂(SO₄)₃ there are from part a. Just think about how many aluminum ions there are per Al₂(SO₄)₃.)
______________ mol

Determine the volume of sulfuric acid solution needed to prepare37.4 g of aluminum sulfate, Al2(SO4)3, by the followingreaction

2Al(s)+3H2SO4 (aq)= Al2(SO4)3 (aq) +3H2 (g)

The sulfuric acid solution, whose density is 1.104 g/ml,contains 15.0% H2SO4 by mass.

2 Al(OH)3(s) + 3 H2SO4(aq) --> Al2(SO4)3(aq) + 6 H2O(l) If 0.500 mol Al(OH)3 and 0.500 mol H2SO4 react, and 48.2 g Al2(SO4)3 are collected from the experiment, what is the percent yield? The molar mass of Al2(SO4)3 is 342.17 g/mol. Be sure to include "%" units with your answer.