Using the balanced equation 3Ba+Al2(SO4)3-->2Al+3BaSO4, how many grams of aluminum will be produced if 4 g of barium react with .45 g of Al2(SO4)3?

Question

Using the balanced equation 3Ba+Al2(SO4)3-->2Al+3BaSO4, how many grams of aluminum will be produced if 4 g of barium react with .45 g of Al2(SO4)3?

Answer 1

We're just worried about the compounds with Al for this question

1. Balance equation (this one is already balanced)

2. Use the grams of Al2(SO4)3 that was given and convert it to moles of Al2(SO4)3

(0.45 g)/(341.96 g) = 0.00132 moles of Al2(SO4)3

**add the molecular weights of all elements in compound to come up with 1 mole = 341.96g Al2(SO4)3**

3. Use the coefficients from balanced equation to convert moles of Al2(SO4)3 to moles of Al

(0.00132 moles of Al2(SO4)3)*(2 moles of Al) = 0.00263 moles of Al

4. Convert moles of Al into grams of Al

(0.00263 moles of Al)*(27 grams of Al) = 0.071 g Al

**use to molecular weight of Al to convert**