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Solution


Q.

When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide produced in grams is _______. (Nearest integer)

Given:
Molar mass of Al is 27.0 g mol-1
Molar mass of O is 16.0 g mol-1                                                          [2025]

Ans.

(153)

Given moles of aluminium =Given mass of AlMolar mass of Al=8127 mol=3 mol

Given moles of oxygen =Given mass of O2Molar mass of O2=12832 mol=4 mol

4Al+3O2⟶2Al2O3

As per stoichiometry of the reaction, 3 mol O2 requires 4 mol Al for complete reaction. So, 4 mol of O2 requires 4×43 mol=5.33 mol Al. But in the reaction only 3 mol of Al is given. So Al is the limiting reagent. Amount of Al2O3 formed depends upon the amount of limiting reagent i.e. Al.  As 4 mol Al gives 2 mol Al2O3. 3 mol of Al will give  1.5 mol of Al2O3. Mass of Al2O3 = number of moles of Al2O3×molar mass of Al2O3=1.5×102 g=153 g.