Question

in the reaction mgcl₂ + 2koh → mg(oh)₂ + 2kcl, if 1 mole mgcl₂ is added to 3 moles koh, what is the limiting reagent?
a. kcl
b. koh
c. mgcl₂
d. mg(oh)₂

Explanation:

Step1: Determine mole - ratio from balanced equation

The balanced equation is $MgCl_2 + 2KOH
ightarrow Mg(OH)_2+2KCl$. The mole - ratio of $MgCl_2$ to $KOH$ is $1:2$.

Step2: Calculate moles of $KOH$ needed for given $MgCl_2$

If we have 1 mole of $MgCl_2$, according to the mole - ratio, the moles of $KOH$ required for complete reaction of $MgCl_2$ is $n_{KOH}=2\times n_{MgCl_2}$. Substituting $n_{MgCl_2} = 1$ mole, we get $n_{KOH}=2\times1 = 2$ moles.

Step3: Compare available and required moles of $KOH$

We have 3 moles of $KOH$ available, and we only need 2 moles of $KOH$ to react completely with 1 mole of $MgCl_2$. So, $MgCl_2$ will be completely consumed first.

Answer:

C. $MgCl_2$