Question

what is the molarity of zncl2 that forms when 20.0 g of zinc completely reacts with cucl2 according to the following reaction? assume a final volume of 285 ml. zn(s) + cucl2(aq) → zncl2(aq) + cu(s)

Explanation:

Step1: Calculate moles of Zn

The molar mass of Zn is 65.38 g/mol. Using the formula $n=\frac{m}{M}$, where $m = 20.0\ g$ and $M=65.38\ g/mol$. So $n_{Zn}=\frac{20.0\ g}{65.38\ g/mol}\approx0.306\ mol$.

Step2: Determine moles of ZnCl₂

From the balanced chemical equation $Zn(s)+CuCl_2(aq)\to ZnCl_2(aq) + Cu(s)$, the mole - ratio of $Zn$ to $ZnCl_2$ is 1:1. So $n_{ZnCl_2}=n_{Zn}=0.306\ mol$.

Step3: Calculate molarity of ZnCl₂

The formula for molarity is $M=\frac{n}{V}$, where $n$ is the number of moles and $V$ is the volume in liters. The volume $V = 285\ mL=0.285\ L$. Then $M_{ZnCl_2}=\frac{0.306\ mol}{0.285\ L}\approx1.07\ M$.

Answer:

1.07