Question

question 5 (1 point) retake question
if 250ml of 0.200 mol/l solution of magnesium chloride is mixed with 250 ml of 0.300 mol/l sodium hydroxide, magnesium hydroxide is a precipitate and the other product, sodium chloride, remains in solution. what is the concentration of sodium chloride solution?
mol/l

Explanation:

Step1: Write the balanced chemical equation

The reaction between magnesium chloride ($\ce{MgCl2}$) and sodium hydroxide ($\ce{NaOH}$) is: $\ce{MgCl2 + 2NaOH -> Mg(OH)2\downarrow + 2NaCl}$.

Step2: Calculate moles of reactants

First, convert volumes to liters. For $\ce{MgCl2}$: volume $V_1 = 250\space mL = 0.250\space L$, concentration $C_1 = 0.200\space mol/L$. Moles of $\ce{MgCl2}$, $n_1 = C_1\times V_1 = 0.200\space mol/L\times0.250\space L = 0.0500\space mol$. For $\ce{NaOH}$: volume $V_2 = 250\space mL = 0.250\space L$, concentration $C_2 = 0.300\space mol/L$. Moles of $\ce{NaOH}$, $n_2 = C_2\times V_2 = 0.300\space mol/L\times0.250\space L = 0.0750\space mol$.

Step3: Determine limiting reactant and moles of $\ce{NaCl}$ produced

From the balanced equation, 1 mole of $\ce{MgCl2}$ reacts with 2 moles of $\ce{NaOH}$. The moles of $\ce{NaOH}$ required to react with 0.0500 mol of $\ce{MgCl2}$ is $2\times0.0500 = 0.100\space mol$, but we only have 0.0750 mol of $\ce{NaOH}$, so $\ce{NaOH}$ is the limiting reactant. From the equation, 2 moles of $\ce{NaOH}$ produce 2 moles of $\ce{NaCl}$, so moles of $\ce{NaCl}$ produced, $n_{NaCl} = n_2 = 0.0750\space mol$ (since 2 moles $\ce{NaOH}$ give 2 moles $\ce{NaCl}$, the mole ratio is 1:1 for $\ce{NaOH}$ and $\ce{NaCl}$ in terms of production here).

Step4: Calculate total volume of the solution

Total volume $V = V_1 + V_2 = 0.250\space L + 0.250\space L = 0.500\space L$.

Step5: Calculate concentration of $\ce{NaCl}$

Concentration of $\ce{NaCl}$, $C_{NaCl} = \frac{n_{NaCl}}{V} = \frac{0.0750\space mol}{0.500\space L} = 0.150\space mol/L$.

Answer:

0.150