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(B) $1$. Calculate moles of reactants:$n_{Al} = \frac{108 \ g}{27 \ g/mol} = 4 \ mol$$n_{MnO} = \frac{213 \ g}{71 \ g/mol} = 3 \ mol$$2$. Determine th

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$102 \ g \ Al_2O_3$ is formed

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(B) $1$. Calculate moles of reactants:$n_{Al} = \frac{108 \ g}{27 \ g/mol} = 4 \ mol$$n_{MnO} = \frac{213 \ g}{71 \ g/mol} = 3 \ mol$$2$. Determine the limiting reagent by comparing $\frac{n}{	ext{stoichiometric coefficient}}$:For $Al$: $\frac{4}{2} = 2$For $MnO$: $\frac{3}{3} = 1$Since $1 $3$. Calculate product formation based on $MnO$:From the reaction,$3 \ mol \ MnO$ produces $1 \ mol \ Al_2O_3$.Therefore,$3 \ mol \ MnO$ produces $1 \ mol \ Al_2O_3$.Mass of $Al_2O_3 = 1 \ mol 	imes (2 	imes 27 + 3 	imes 16) \ g/mol = 1 	imes 102 \ g = 102 \ g$.Thus,option $B$ is correct.

$108 \ g \ Al$ and $213 \ g \ MnO$ were heated to initiate the reaction:
$2 Al + 3 MnO \to Al_2O_3 + 3 Mn$
Which of the following is correct? (Atomic weights: $Mn = 55$,$Al = 27$)

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$108 \ g \ Al$ and $213 \ g \ MnO$ were heated to initiate the reaction:$2 Al + 3 MnO \to Al_2O_3 + 3 Mn$Which of the following is correct? (Atomic weights: $Mn = 55$,$Al = 27$)