Chapter 7, Problem 7.90QA

Interpretation Introduction

To find:

a) How much chalcopyrite had to be mined to produce one dollar’s worth of these pennies?

b) How much chalcopyrite had to be mined to produce one dollar’s worth of the pennies if the first reaction had a percent yield of $85 \%$ and the second and third reactions had percent yields of essentially $100 \%?$

c) How much chalcopyrite had to be mined to produce one dollar’s worth of the pennies if each of the reactions proceeded with $85 \%$ yield?

Answer to Problem 7.90QA

Solution:

a) $852 g$ chalcopyrite had to be mined to produce one dollar’s worth of the pennies.

b) $1004 g$ chalcopyrite had to be mined to produce one dollar’s worth of the pennies if the first reaction had a percent yield of $85 \%$ and the second and third reactions had percent yields of essentially $100 \%.$

c) $1388\mathrm{ }\mathrm{g}$ chalcopyrite had to be mined to produce one dollar’s worth of the pennies if each of the reactions proceeded with $85 \%$ yield.

Explanation of Solution

a) To find amount of chalcopyrite that had to be mined to produce one dollar’s worth of these pennies.

1) Formula and concept:

We are given mass of one penny, that is, $3.11\mathrm{ }\mathrm{g}$ containing $95\mathrm{ }\mathrm{\%}\mathrm{ }\mathrm{C}\mathrm{u}\mathrm{ }\mathrm{b}\mathrm{y}\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}$. So first, we have to find the mass of Cu in 1 penny.

$\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{C}\mathrm{u}=\mathrm{ }\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{C}\mathrm{u}}{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }1\mathrm{ }\mathrm{p}\mathrm{e}\mathrm{n}\mathrm{n}\mathrm{y}}\mathrm{ }\times 100$

Then we know that $1 dollar = 100 pennies$. So we have to find the amount of $Cu$ for 100 pennies.

The reactions involved in this process of recovering $Cu$ metal from $CuFe{S}_2$ are

$2 CuFe{S}_{2 \left(s\right)}+3 O_{2 \left(g\right)}\to 2 Cu{S}_{ \left(s\right)}+2 Fe{O}_{ \left(s\right)}+2 S{O}_{2 \left(g\right)}$

$2 Cu{S}_{ \left(s\right)}\to C{u}_2{S}_{ \left(s\right)}+S_{ \left(s\right)}$

$C{u}_2{S}_{ \left(s\right)}+S+O_{2 \left(g\right)}\to 2 C{u}_{ \left(s\right)}+2 S{O}_{2 \left(g\right)}$

Using the mass of $Cu$ and molar ratios from the balanced reactions, we can determine the mass of  $CuFe{S}_2$ required.

2) Given:

We are given mass of one penny, that is, $3.11 g$ containing $95 \% Cu\mathrm{ }$ by mass.

3) Calculations:

The mass of Cu metal present in one penny is given by

$\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{a}\mathrm{g}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{C}\mathrm{u}=\mathrm{ }\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{C}\mathrm{u}}{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }1\mathrm{ }\mathrm{p}\mathrm{e}\mathrm{n}\mathrm{n}\mathrm{y}}\mathrm{ }\times 100$

$95\mathrm{ }\mathrm{\%}=\mathrm{ }\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{C}\mathrm{u}}{3.11\mathrm{ }\mathrm{g}}\mathrm{ }\times 100$

$\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{C}\mathrm{u}=2.95\mathrm{ }\mathrm{g}$

So, the mass of Cu for 1 dollar is $=2.95\mathrm{ }\mathrm{g}\times 100=295.45\mathrm{ }\mathrm{g}\mathrm{ }\mathrm{C}\mathrm{u}$

To find mass of $CuFe{S}_2$ required to get $295 g Cu$: We are getting $2\mathrm{ }\mathrm{C}\mathrm{u}$ from 1 $C{u}_2{S}_{ }$,  1 $C{u}_2{S}_{ }$from $2<\mathrm{i}\mathrm{n}\mathrm{l}\mathrm{i}\mathrm{n}\mathrm{e}-\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{a}><\mathrm{m}:\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{h}><\mathrm{m}:\mathrm{m}\mathrm{i}>\mathrm{C}</\mathrm{m}:\mathrm{m}\mathrm{i}><\mathrm{m}:\mathrm{m}\mathrm{i}>\mathrm{u}</\mathrm{m}:\mathrm{m}\mathrm{i}><\mathrm{m}:\mathrm{m}\mathrm{i}>\mathrm{S}</\mathrm{m}:\mathrm{m}\mathrm{i}></\mathrm{m}:\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{h}></\mathrm{i}\mathrm{n}\mathrm{l}\mathrm{i}\mathrm{n}\mathrm{e}-\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{a}>$, $2$$\mathrm{C}\mathrm{u}\mathrm{S}$ from $2\mathrm{ }\mathrm{C}\mathrm{u}\mathrm{F}\mathrm{e}{\mathrm{S}}_2,$ so we are getting $2\mathrm{ }\mathrm{C}\mathrm{u}$ from $2\mathrm{ }\mathrm{C}\mathrm{u}\mathrm{F}\mathrm{e}{\mathrm{S}}_2$. They are in a one to one molar ratio.

$295.45 g Cu\times \frac{1 mol Cu}{63.546 g}\times \frac{1 mol CuFe{S}_2 }{1 mol Cu}\times \frac{183.52 g}{1 mol CuFe{S}_2}=853 g CuFe{S}_2$

So the amount of $CuFe{S}_2$ required is $852 g$, assuming $100\%$ yield.

b) To find the amount of chalcopyrite that had to be mined to produce one dollar’s worth of these pennies if the first reaction had percent yield of $85\mathrm{ }\mathrm{\%}$ and other two reactions gives $100\mathrm{ }\mathrm{\%}$ yield.

1) Formula and concept:

The first reaction has $85\mathrm{ }\mathrm{\%}$  yield of $CuS.$ So from mass of $Cu$ given, we have to find mass of $CuS$, that is, the mass of $\mathrm{C}\mathrm{u}\mathrm{S}$ needed to produce 295 g $Cu$. From that, we have to determine mass of $\mathrm{C}\mathrm{u}\mathrm{S}$ that could have been produced with $100\mathrm{ }\mathrm{\%}$ yield. Using molar ratios, we have to find mass of $CuFe{S}_2$ needed to produce this much of $CuS$.

2) Calculations:

The amount of $C{u}_{2}S$ for $295.45g Cu$(assuming $100\%$ yield) for the third reaction is

$295.45 g Cu\times \frac{1 mol Cu}{63.546 g}\times \frac{1 moles CuS }{2 moles Cu}\times \frac{159.157 g}{1 mol CuS}=369.99 g {Cu}_{2}S$

$369.99\mathrm{g}\mathrm{ }{\mathrm{C}\mathrm{u}}_{2}S$ is $100\%\mathrm{ }$ yield for the second reaction. So, calculating the amount of $CuS$ required to produce $369.99 g C{u}_{2}S$ as

$369.99 g {Cu}_{2}S\times \frac{1 mol {Cu}_{2}S}{159.157 g}\times \frac{2 mol CuS }{1mol {Cu}_{2}S}\times \frac{95.61 g}{1 mol CuS}=444.52 g CuS$

Now, for the third reaction, the percent yield is 85%. So, calculating the theoretical yield as

$Percent\mathrm{ }yield=\mathrm{ }\frac{Actual\mathrm{ }yield}{Theoretical\mathrm{ }yield}\mathrm{ }\times 100$

$85\mathrm{ }\mathrm{\%}=\mathrm{ }\frac{444.52\mathrm{ }g\mathrm{ }CuS}{Theoretical\mathrm{ }yield}\mathrm{ }\times 100$

So, the theoretical yield is $522.97 g CuS$. Now, calculating amount of $CuFe{S}_2$ as

$522.97 g CuS\times \frac{1 mol CuS}{95.611 g}\times \frac{1 mole CuFe{S}_2}{1 mole CuS}\times \frac{183.54 g}{1 mol CuFe{S}_2}=1004 g CuFe{S}_2$

So the amount of $CuFe{S}_2$ needed is $1004\mathrm{ }g$.

c) To find the amount of chalcopyrite that had to be mined to produce one dollar’s worth of the pennies if each of the reactions proceeded with 85 % yield.

We have to follow the same procedure as above, considering each reaction gives $85\mathrm{ }\mathrm{\%}$ practical yield. We have to calculate theoretical yields for each reactions separately to get the amount of $CuFe{S}_2$ required.

The amount of Cu that could be produced is

$Percent\mathrm{ }yield=\mathrm{ }\frac{Actual\mathrm{ }yield}{Theoretical\mathrm{ }yield}\mathrm{ }\times 100$

$85\mathrm{ }\mathrm{\%}=\mathrm{ }\frac{295.45\mathrm{ }g\mathrm{ }Cu}{Theoretical\mathrm{ }yield}\mathrm{ }\times 100$

$Theoretical\mathrm{ }yield\mathrm{ }of\mathrm{ }Cu=347.588\mathrm{ }g$

i. The amount of $C{u}_2{S}_{\mathrm{ }}$  used to get $347.588\mathrm{ }g\mathrm{ }Cu$ is

$347\mathrm{ }g\mathrm{ }Cu\times \frac{1\mathrm{ }mol\mathrm{ }Cu}{63.54\mathrm{ }g}\times \frac{1\mathrm{ }mole\mathrm{ }C{u}_2{S}_{\mathrm{ }}}{2\mathrm{ }moles\mathrm{ }Cu}\times \frac{159.157\mathrm{ }g}{1\mathrm{ }mol\mathrm{ }C{u}_2{S}_{\mathrm{ }}}=435.28\mathrm{ }g\mathrm{ }C{u}_2{S}_{\mathrm{ }}$

The amount of $C{u}_2{S}_{\mathrm{ }}$ that could have been made is

$85\mathrm{ }\mathrm{\%}=\mathrm{ }\frac{435\mathrm{ }g\mathrm{ }C{u}_2{S}_{\mathrm{ }}}{Theoretical\mathrm{ }yield\mathrm{ }of\mathrm{ }C{u}_2{S}_{\mathrm{ }}}\mathrm{ }\times 100$

$Theoretical\mathrm{ }yield\mathrm{ }of\mathrm{ }C{u}_2{S}_{\mathrm{ }}=512\mathrm{ }g$.

The amount of  $CuS$ used to get $512\mathrm{ }g\mathrm{ }C{u}_2{S}_{\mathrm{ }}$ is

$512\mathrm{ }g\mathrm{ }C{u}_2{S}_{\mathrm{ }}\times \frac{1\mathrm{ }mol\mathrm{ }C{u}_2{S}_{\mathrm{ }}}{159.157\mathrm{ }g}\times \frac{2\mathrm{ }moles\mathrm{ }CuS}{1\mathrm{ }mol\mathrm{ }C{u}_2{S}_{\mathrm{ }}}\times \frac{95.611\mathrm{ }g}{1\mathrm{ }mol\mathrm{ }CuS}=615\mathrm{ }g\mathrm{ }CuS$

The amount of $CuS$ that could have been made is

$85\mathrm{ }\mathrm{\%}=\mathrm{ }\frac{615\mathrm{ }g\mathrm{ }CuS}{Theoretical\mathrm{ }yield\mathrm{ }of\mathrm{ }CuS}\mathrm{ }\times 100$

$Theoretical\mathrm{ }yield\mathrm{ }of\mathrm{ }CuS=723\mathrm{ }g$

The amount of  $CuFe{S}_2$ used to get $723\mathrm{ }g\mathrm{ }CuS$ is

$723 g CuS\times \frac{1 mol CuS}{95.611 g}\times \frac{1 mol CuFe{S}_2}{2 mol CuS}\times \frac{183.54 g}{1 mol CuFe{S}_2}=1388 g CuFe{S}_2$

So, $1388\mathrm{ }\mathrm{g}\mathrm{ }\mathrm{C}\mathrm{u}\mathrm{F}\mathrm{e}{\mathrm{S}}_2$ had to be mined to produce one dollar’s worth of pennies.

Conclusion:

From the balanced equations, we can find out quantitative information about the reactants that need to be used or the products that could form.