we wish to determine the moles of mg(oh)₂ produced when 250 ml of 2.0 m koh reacts with excess mg(no₃)₂…

we wish to determine the moles of mg(oh)₂ produced when 250 ml of 2.0 m koh reacts with excess mg(no₃)₂. mg(no₃)₂(aq) + 2koh(aq) → 2kno₃(aq) + mg(oh)₂(s) in the previous step, you determined 0.50 mol koh react. how many moles of mg(oh)₂ form during the reaction?

we wish to determine the moles of mg(oh)₂ produced when 250 ml of 2.0 m koh reacts with excess mg(no₃)₂. mg(no₃)₂(aq) + 2koh(aq) → 2kno₃(aq) + mg(oh)₂(s) in the previous step, you determined 0.50 mol koh react. how many moles of mg(oh)₂ form during the reaction?

Answer

Explanation:

Step1: Identify mole ratio

From the reaction: $\ce{Mg(NO_{3}){2}(aq) + 2KOH(aq) -> 2KNO{3}(aq) + Mg(OH){2}(s)}$, the mole ratio of $\ce{KOH}$ to $\ce{Mg(OH){2}}$ is $2:1$.

Step2: Calculate moles of $\ce{Mg(OH)_{2}}$

Given moles of $\ce{KOH} = 0.50\ \text{mol}$. Let moles of $\ce{Mg(OH){2}}$ be $x$. Using the ratio $\frac{\text{moles of } \ce{KOH}}{\text{moles of } \ce{Mg(OH){2}}} = \frac{2}{1}$, we have $\frac{0.50}{x} = \frac{2}{1}$. Solving for $x$: $x = \frac{0.50}{2} = 0.25\ \text{mol}$.

Answer:

$0.25$ moles