Question

A sample is a mixture of AgNO₃, CuCl₂, and FeCl₃.When a 1.0000-g sample of the mixture is dissolved inwater and reacted with excess silver nitrate, 1.7809 g ofprecipitate forms. When a separate 1.0000-g sample ofthe mixture is treated with a reducing agent, all themetal ions in the mixture are reduced to pure metals.The total mass of pure metals produced is 0.4684 g.Calculate the mass percent of AgNO₃, CuCl₂, and FeCl₃in the original mixture.

Short answer

The mass percent of AgNO₃, CuCl₂, and FeCl₃in the original mixture is 25%, 40% and 35% respectively.

Step-by-step solution

Step 1:Determinethe mass of different elements

$x+y+z=1.000g$Let x be the mass of AgNO₃, y be the mass of CuCl₂ and z be the mass of FeCl_3.

Therefore, we can write .

The present in CuCl₂and FeCl₃ reacts with excess and forms precipitate of AgCl (s).

Atomic mass of AgCl is .$107.9$

The mass of Cl present in the mixture and in different elements have been calculated below.

Mass of Cl in mixture

$\begin{array}{l}=1.7809gAgCl\times \frac{35.45gCl}{143.35gAgCl}\\ =0.4404gCl\end{array}$

Mass of Cl in CuCl₂

$\begin{array}{l}=zgFeC{l}_3\times \frac{3\left(35.45\right)gCl}{162.20gFeC{l}_3}\\ =0.5273y\end{array}$

Mass of Cl in FeCl₃

$\begin{array}{l}=zgFeC{l}_3\times \frac{3\left(35.45\right)gCl}{162.20gFeC{l}_3}\\ =0.5273y\end{array}$

Therefore, we can write

$0.4404gCl=0.5273y+0.6557z$

Now the mass of metals in each salt must be calculated.

Mass of Ag in AgNO₃

$\begin{array}{l}=xgAgN{O}_3\times \frac{107.9gAg}{169.91gAgN{O}_3}\\ =0.6350x\end{array}$

Mass of Cu in CuCl₂

$\begin{array}{l}=y-0.5273y\\ =0.4727y\end{array}$

Mass of Fe in FeCl₃

$\begin{array}{l}=z-0.6557z\\ =0.3443z\end{array}$

Therefore, we can write

$0.4684gmetals=0.6350x+0.4727y+0.3443z$

Determine the mass percent of AgNO3, CuCl2, and FeCl3 in the original mixture

Now we have three equations with three unknows

$\begin{array}{c}x+y+z=1\\ 0.5273y+0.6557z=0.4404\\ 0.6350x+0.4727y+0.3443z=0.4684\end{array}$

Solving the above three equations, we get

$\begin{array}{l}z=0.3494gFeC{l}_3\\ y=0.4007gCuC{l}_2\\ x=0.2499gAg\end{array}$

The mass percent of AgNO₃ will be

$\begin{array}{l}\frac{0.2499g}{1.000g}\times 100\\ =24.99\%\text{\hspace{0.17em}\hspace{0.17em}}AgN{O}_3\\ \approx 25\%AgN{O}_3\end{array}$

The mass percent of CuCl₂ will be

$\begin{array}{l}\frac{0.4007g}{1.000g}\times 100\\ =40.07\%\text{\hspace{0.17em}\hspace{0.17em}}CuC{l}_2\\ \approx 40\%CuC{l}_2\end{array}$

The mass percent of FeCl₃ will be

$\begin{array}{l}\frac{0.3494g}{1.000g}\times 100\\ =34.94\%\text{\hspace{0.17em}\hspace{0.17em}}FeC{l}_3\\ \approx 35\%FeC{l}_3\end{array}$