When 81.0 g of aluminium is allowed to react | vedclass

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Question

(C) The balanced chemical equation is: $4Al + 3O_2 \longrightarrow 2Al_2O_3$. Moles of $Al = \frac{81.0 \ g}{27.0 \ g \ mol^{-1}} = 3.0 \ mol$. Moles

Vedclass · Accepted Answer

$153$

Vedclass · Answer

(C) The balanced chemical equation is: $4Al + 3O_2 \longrightarrow 2Al_2O_3$. Moles of $Al = \frac{81.0 \ g}{27.0 \ g \ mol^{-1}} = 3.0 \ mol$. Moles of $O_2 = \frac{128.0 \ g}{32.0 \ g \ mol^{-1}} = 4.0 \ mol$. According to the stoichiometry,$4 \ mol$ of $Al$ requires $3 \ mol$ of $O_2$. For $3.0 \ mol$ of $Al$,required $O_2 = \frac{3}{4} 	imes 3.0 = 2.25 \ mol$. Since $4.0 \ mol$ of $O_2$ is available,$Al$ is the limiting reagent. Moles of $Al_2O_3$ produced $= \frac{2}{4} 	imes 3.0 = 1.5 \ mol$. Molar mass of $Al_2O_3 = (2 	imes 27.0) + (3 	imes 16.0) = 54.0 + 48.0 = 102.0 \ g \ mol^{-1}$. Mass of $Al_2O_3 = 1.5 \ mol 	imes 102.0 \ g \ mol^{-1} = 153.0 \ g$.

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 $16.0 \ g \ mol^{-1}$.

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