Question

In the reaction,
$2Mg+O_2\to 2MgO$
Which of the following is true when 4.8 g of $Mg$ and 6.4 g of $O_2$ react?

• A. All the oxygen will be consumed
• B. $0.2$ mol of $MgO$ will be produced
• C. $0.5$ mol of $Mg$ will remain unreacted
• D. All the above

Answer 1

The correct option is B $0.2$ mol of $MgO$ will be produced

We know, molar mass of $Mg$ = 24 g
$\therefore$ 4.8 g $Mg$ = $\frac{4.8}{24}$ mol of $Mg=0.2mol$
Again molar mass of of $O_2$ = 32 g
$\therefore$ 6.4 g $O_2$ = $\frac{6.4}{32}$ mol of $O_2=0.2mol$
Given:
$2Mg+O_2\to 2MgO$
2 moles of $Mg$ react with 1 mole of $O_2$
So, 0.2 mole of $Mg$ will react with 0.1 moles of $O_2$
Hence, $Mg$ is the limiting reagent here.

From the reaction, 2 moles of $Mg$ gives 2 moles of $MgO$
So, 0.2 mole of $Mg$ will give 0.2 mole of $MgO$