5 mol of aluminium chloride is mixed with 280 g of magnesium oxide, to give aluminium oxide. Find the amount of the excess reagent left after the reaction. 2AlCl3+3MgO→Al2O3+3MgCl2
A
44.055 g
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B
42.032 g
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C
54.056 g
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D
32.056 g
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Solution
The correct option is A 44.055 g 2AlCl3+3MgO→Al2O3+3MgCl2 Moles of MgO=28040=7 mol of MgO 2 mol of AlCl3 reacts with 3 mol of MgO 5 mol will react with = 7.5 mol So, MgO is the limiting reactant. 3 mol of MgO requires 2 mol of AlCl3 7 mol of MgO will require = 4.67 mol Excess reagent left = 5 – 4.67 = 0.33 mol Mass of AlCl3 left =0.33×133.5=44.055 g