Question

What will be a, b and c for the following reaction to be balanced?

$aFeC{l}_3+bMgO\to cF{e}_2{O}_3+dMgC{l}_2$

• A. a = $2$, b = $3$, c = $3$, d = $4$
• B. a = $2$, b = $4$, c = $3$, d = $3$
• C. a = $1$, b = $\frac{3}{2}$, c = $\frac{1}{2}$, d = $2$
• D. a = $1$, b = $\frac{3}{2}$, c = $\frac{1}{2}$, d = $\frac{3}{2}$

Answer 1

The correct option is D

a = $1$, b = $\frac{3}{2}$, c = $\frac{1}{2}$, d = $\frac{3}{2}$

$aFeC{l}_3+bMgO\to cF{e}_2{O}_3+dMgC{l}_2$

Now, for iron we will equate the number of atoms on the reactant and product sides: a = 2c

Chlorine: 3a = 2d

Magnesium: b = d

Oxygen: b = 3c

Now we take a = 1 and put this in a = 2c which gives C = $\frac{1}{2}$ and from 3a = 2d we will get D = $\frac{3}{2}$ and from b = d we get b = $\frac{3}{2}$

So, the balanced equation is:
$FeC{l}_3+\frac{3}{2}MgO\to \frac{1}{2}F{e}_2{O}_3+\frac{3}{2}MgC{l}_2$