Question

b. calculate the mass (in g) of cl₂(g) produced if 18.2 grams hcl(aq) of react with excess mno₂(s).

Explanation:

Step1: Balance the chemical equation

$$4HCl(aq)+MnO_2(s) ightarrow MnCl_2(aq)+2H_2O(g)+Cl_2(g)$$

Step2: Calculate the molar mass of HCl

The molar mass of HCl ($M_{HCl}$) is $M_{HCl}=1 + 35.5=36.5\ g/mol$

Step3: Calculate the number of moles of HCl

Given mass of HCl ($m_{HCl}$) = 18.2 g. Using the formula $n=\frac{m}{M}$, the number of moles of HCl ($n_{HCl}$) is $n_{HCl}=\frac{18.2\ g}{36.5\ g/mol}=0.5\ mol$

Step4: Use the mole - ratio from the balanced equation

From the balanced equation $4HCl(aq)+MnO_2(s)
ightarrow MnCl_2(aq)+2H_2O(g)+Cl_2(g)$, the mole ratio of $HCl$ to $Cl_2$ is $n_{HCl}:n_{Cl_2}=4:1$. Let the number of moles of $Cl_2$ be $n_{Cl_2}$. Then $n_{Cl_2}=\frac{n_{HCl}}{4}$
Substituting $n_{HCl} = 0.5\ mol$, we get $n_{Cl_2}=\frac{0.5\ mol}{4}=0.125\ mol$

Step5: Calculate the molar mass of $Cl_2$

The molar mass of $Cl_2$ ($M_{Cl_2}$) is $M_{Cl_2}=2\times35.5 = 71\ g/mol$

Step6: Calculate the mass of $Cl_2$

Using the formula $m = n\times M$, the mass of $Cl_2$ ($m_{Cl_2}$) is $m_{Cl_2}=n_{Cl_2}\times M_{Cl_2}$
Substituting $n_{Cl_2}=0.125\ mol$ and $M_{Cl_2}=71\ g/mol$, we get $m_{Cl_2}=0.125\ mol\times71\ g/mol = 8.875\ g$

Answer:

$8.875\ g$