If 11.7 g of aluminum reacts with 37.2 g of copper (ii) sulfate according to the following reaction, how many grams of aluminum sulfate will be produced?

There is 26.59 grams of aluminium sulfate produced

Explanation:

Step 1: Data given

Mass of aluminium = 11.7 grams

Mass of copper (II) sulfate = 37.2 grams

Molar mass of Aluminium = 26.98 g/mol

Molar mass of CuSO4 = 159.61 g/mol

Molar mass of Al2(SO4)3 = 342.15 g/mol

Step 2: The balanced equation

2Al + 3CuSO4 → Al2(SO4)3 + 3Cu

Step 3: Calculate moles of Aluminium

Moles Al = mass Al / molar mass Al

Moles Al = 11.7 grams / 26.98 g/mol

Moles Al = 0.434 mol

Step 4: Calculate moles of CuSO4

Moles CuSO4 = 37.2 grams / 159.61 g/mol

Moles CuSO4 = 0.233 moles

Step 5: Calculate limiting reactant

For 2 moles of Al we need 3 moles of CuSO4

CuSO4 is the limiting reactant. It will be completely consumed (0.233 moles).

Al is in excess. There will be consumed 0.233 *(2/3) = 0.1553 moles

There will remain 0.434 - 0.1553 = 0.2787 moles

Step 6: Calculate moles of Al2(SO4)3

For 2 moles of Al we need 3 moles of CuSO4, to produce 1 mole of Al2(SO4)3 and  3 moles of Cu

For 0.233 moles CuSO4 we produce 0.233/3 = 0.0777 moles of Al2(SO4)3

Step 7: Calculate mass of Al2(SO4)3

Mass of Al2(SO4)3 = moles Al2(SO4)3 * molar mass Al2(SO4)3

Mass of Al2(SO4)3 = 0.0777 moles * 342.15g/mol

Mass of Al2(SO4)3 = 26.59 grams

There is 26.59 grams of aluminium sulfate produced

the answer is b. dilution

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