Question

The following chemical equation represents a reaction:
$Mg\left(s\right)+2HCl\left(aq\right)\to MgC{l}_2\left(aq\right)+H_2\left(g\right)$
How many grams of the excess reactant will be left over in the reaction when $6g$ of $HCl$ is made to react with $5g$ of $Mg$?

• A. $3.03g$
• B. $2.05g$
• C. $4.02g$
• D. $1.25g$

Answer 1

The correct option is A $3.03g$

According to the given reaction,
1 mole of $Mg$ reacts with 2 moles of $HCl$.
24 g of $Mg$ reacts with $(2\times 36.5)$ g of $HCl$.
5 g of $Mg$ will react with $\frac{2\times 36.5}{24}\times 5=15.20gofHCl$
But, there is only 6 g of $HCl$ available to react with the 5 g of $Mg$. (given in the question)
Therefore, $HCl$ will be the limiting reagent.
Similarly, $(2\times 36.5)$ g of $HCl$ reacts with 24 g of $Mg$.
6 g of $HCl$ reacts with $\frac{24}{2\times 36.5}\times 6=1.97$ g
Excess $Mg$ = 5.00 - 1.97 g = 3.03 g