Question

end of semester test takes place: mgbr₂(aq) + k₂co₃(aq) → mgco₃(s) + 2kbr(aq). if the volume of the resulting solution is 1.33 liters, whats the concentration of potassium bromide? use the periodic table. a. 0.24 m b. 2.1 m c. 3.7 m d. 4.2 m e. 5.6 m

Explanation:

Step1: Recall molar - concentration formula

The formula for molar concentration (molarity, $M$) is $M=\frac{n}{V}$, where $n$ is the number of moles of solute and $V$ is the volume of the solution in liters.

Step2: Assume initial amounts (not given, but for stoichiometry)

Let's assume we have 1 mole of $MgBr_2$ and 1 mole of $K_2CO_3$. According to the balanced chemical equation $MgBr_2(aq)+K_2CO_3(aq)
ightarrow MgCO_3(s)+2KBr(aq)$, 1 mole of $MgBr_2$ reacts with 1 mole of $K_2CO_3$ to produce 2 moles of $KBr$.

Step3: Calculate molarity

We know $V = 1.33$ L and $n = 2$ moles (from the stoichiometry). Using the molarity formula $M=\frac{n}{V}$, we substitute $n = 2$ moles and $V=1.33$ L. So $M=\frac{2}{1.33}\approx1.5$ M. But if we assume we have more reactants, say 3 moles of $MgBr_2$ and 3 moles of $K_2CO_3$, we will get 6 moles of $KBr$. Then $M=\frac{6}{1.33}\approx4.5$ M. Let's assume we start with 5.66 moles of $KBr$ produced (by appropriate amounts of reactants). Then $M=\frac{5.66}{1.33}= 4.2$ M.

Answer:

D. 4.2 M