Question

$2Al+3MnO\stackrel{\Delta }{\to }A{l}_2{O}_3+3Mn$
$108.0$ g of $Al$ and $213.0$ g of $MnO$ were heated to initiate the reaction. (Molecular weight of $MnO=71$ g/mol and atomic weight of $Al=13$ g)

Which of the following statements is/are correct?

• A. $Al$ is present in excess.
• B. $MnO$ is present is excess.
• C. $54.0$ g of $Al$ is required.
• D. $159.0$ g of $MnO$ is in excess.

Answer 1

Answer: (A), (C)

The correct options are
A $Al$ is present in excess.
C $54.0$ g of $Al$ is required.

Moles of $Al=\frac{108}{27}=4.0$ mol
Moles of $MnO=\frac{213}{71}=3.0$ mol
Since $2$ mol of $Al$ requires $3$ mol of $MnO$, $Al$ is in excess.
Weight of $Al$ required $=(3\times \frac{2}{3}\times 27)=54.0$ g of $Al$

Weight of $Al$ in excess $=(108-54)=54.0$ g