Question

How much magnesium sulphide can be obtained from $2.00$ g of $Mg$ and $2.00$ g of $S$ by the following reaction.

$Mg+S\to MgS$.

• A. $3.5g$
• B. $4g$
• C. $2.5g$
• D. $4.5g$

Answer 1

The correct option is A $3.5g$

$\underset{24g}{Mg}+\underset{32g}{S}\to \underset{24+32=56g}{MgS}$

32 g of S in obtained from 24 g of Mg.

2 g of S is obtained from $=\frac{24}{32}\times 2=1.5g$ of Mg

So, S is completely consumed, and Mg is left.

Hence, S is the limiting reagent.

Therefore, the amount of product, i.e., MgS, will be

determined from S and not from Mg.

32g of $S=56g$ of MgS

2 g of $S=\frac{56}{32}\times 2=3.5g$ of MgS