When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide

Question

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 $27.0 \mathrm{~g} \mathrm{~mol}^{-1}$\ Molar mass of O is $16.0 \mathrm{~g} \mathrm{~mol}^{-1}$\

Professor PC Thomas Classes · Answer

(153) : The balanced chemical reaction can be written as :\ $4 \mathrm{Al}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}$\ Moles of $\mathrm{Al}=\frac{81}{27}=3$ moles\ Moles of oxygen $=\frac{128}{32}=4$ moles\ From the given chemical reaction, it is clear that Al is limiting reagent.\ $	herefore \quad$ Moles of $\mathrm{Al}_{2} \mathrm{O}_{3}$ formed $=\frac{1}{2} 	imes 3 \mathrm{~mol}$\ Mass of $\mathrm{Al}_{2} \mathrm{O}_{3}$ formed $=\frac{3}{2} 	imes 102=153 \mathrm{~g}$\