Answer:
There is 26.59 grams of aluminium sulfate produced
Explanation:
<u>Step 1:</u> 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
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<u>Step 2</u>: The balanced equation
2Al + 3CuSO4 → Al2(SO4)3 + 3Cu
<u>Step 3</u>: 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
<u>Step 4</u>: Calculate moles of CuSO4
Moles CuSO4 = 37.2 grams / 159.61 g/mol
Moles CuSO4 = 0.233 moles
<u>Step 5:</u> 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
<u>Step 6: </u>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
<u>Step 7</u>: 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
