Question

Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide: 4HCl(aq)+MnO2(s) ---> MnCl2(aq)+2H2O(l)+Cl2(g) ?

You add 42.7 g of MnO2 to a solution containing 47.1 g of HCl.

(a). What is the limiting reactant?

(b). What is the theoretical yield of Cl2?

(c). If the yield of the reaction is 79.1%, what is the actual yield of chlorine?

Answer 1

a) HCl
b) 22.9g
c) 18.11g

Explanation:

Hey!

So let's start by calculating the mols of each substance using a periodic table.

MMn = 54.94g/mol
MO2 = 2(16) = 32g/mol
MH = 1g/mol
MCl = 35.45g/mol

Molar Mass of MnO2:
`54.94+2(16)=86.94`

Molar Mass of HCl:
`1+35.45=36.45`

Looking at the equation, we need a 4:1 ratio of HCl to MnO2 to make the reaction work, let's see if we have four times as much of the HCl.

Mols of MnO2:
`42.7/86.94=0.49`

Mols of HCl:
`47.1/36.45=1.29`

As we can see, we have less than the amount of HCl we need for all of the MnO2 to react, therefore the HCl is the limiting reactant.

For the theoretical yield of Cl2, we need to determine how many mols will be produced. As we can see by the equation, four mols of HCl make one mol of Cl2.

Molar Mass of Cl2:
`35.45*2=70.9`

Since we have 1.29mol of HCl, and we know it is the limiting reactant:

Mols of Cl2
`1.29/4=0.323`

Mass of Cl2 (Theoretical yield)
`0.323*70.9=22.9`

To calculate the actual yield, we multiply the theoretical yield by the final percentage:

`22.9*0.791=18.11`

Therefore, the actual yield of chlorine is 18.11g.

Good luck!

Answer 2

`(a)` The limiting reactant is `"HCl"` because it produces less `"Cl"_2"`.

`(b)` The theoretical yield of `"Cl"_2"` is `"22.9 g"`.

`(c)` The actual yield is `"18.1 g Cl"_2"`.

Explanation:

Balanced equation

`"4HCl(aq) + MnO"_2("s")` `rarr` `"MnCl"_2("aq") + "2H"_2"O("l") + Cl"_2("g")`

Molar masses

`M_"HCl"=(1xx"1.008 g/mol H") + (1xx"35.453g/mol Cl")="36.461 g/mol"`

`M_"MnO2"=(1xx"54.938 g/mol Mn") + (2xx"15.999 g/mol O")="86.936 g/mol"`

`M_"Cl2"=(2xx"35.453 g/mol Cl"_2")"="70.906 g/mol"`

There are three basic steps to go from mass of the reactant to mass of `"Cl"_2"`.

`color(red)"mass reactant"` `rarr` `color(red)("mol reactant"`

1. Multiply the given mass of the reactant by the reciprocal of its molar mass (mol/g). This will give you moles of reactant. This is the same as dividing the given mass by the molar mass.

`color(red)("mol reactant"` `rarr` `color(blue)"mol Cl"_2"`

1. Multiply mol reactant by the mol ratio between the reactant and `"Cl"_2"` from the balanced equation, with `"Cl"_2"` in the numerator.

`color(blue)"mol Cl"_2"` `rarr` `color(green)"mass Cl"_2"`

1. Multiply mol `"Cl"_2"` by its molar mass. This will give you mass of `"Cl"_2"`.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(red)"mass MnO"_2"` `rarr` `color(red)("mol MnO"_2"`

`42.7color(red)cancel(color(black)("g MnO"_2))xx(1"mol MnO"_2)/(86.936color(red)cancel(color(black)("g MnO"_2)))="0.49117 mol MnO"_2"`

I am keeping a couple of guard digits to reduce rounding errors. I will round the mass of chlorine gas to three significant figures.

`color(red)"mol MnO"_2"` `rarr` `color(blue)("mol Cl"_2"`

`0.49117color(red)cancel(color(black)("mol MnO"_2))xx(1"mol Cl"_2)/(1color(red)cancel(color(black)("mol MnO"_2)))="0.49117 mol Cl"_2"`

`color(blue)("mol Cl"_2"` `rarr` `color(green)("mass Cl"_2"`

`0.49117color(red)cancel(color(black)("mol Cl"_2))xx(70.906"g Cl"_2)/(1color(red)cancel(color(black)("mol Cl"_2)))="34.8 g Cl"_2"` (rounded to three significant figures)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(red)("mass HCl"` `rarr` `color(red)("mol HCl"`

`47.1color(red)cancel(color(black)("g HCl"))xx(1"mol HCl")/(36.461color(red)cancel(color(black)("g HCl")))="1.2918 mol HCl"`

`color(red)"mol HCl"` `rarr` `color(blue)("mol Cl"_2"`

`1.2918 color(red)cancel(color(black)("mol HCl"))xx(1"mol Cl"_2)/(4color(red)cancel(color(black)("mol HCl")))="0.32295 mol Cl"_2"`

`color(blue)("mol Cl"_2"` `rarr` `color(green)("mass Cl"_2"`

`0.32295color(red)cancel(color(black)("mol Cl"_2))xx(70.906"g Cl"_2)/(1color(red)cancel(color(black)("mol Cl"_2)))="22.9 g Cl"_2"` (rounded to three significant figures)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Questions

`(a)` The limiting reactant is `"HCl"` because it produces less `"Cl"_2"`.

`(b)` The theoretical yield of `"Cl"_2"` is `"22.9 g"`.

`(c)` `"% yield"=("actual yield")/("theoretical yield")xx100%"`

`79.1%=("actual yield")/("22.9 g")xx100%`

Divide both sides by `100%`.

`0.791=("actual yield")/("22.9 g")`

Multiply both sides by `"22.9 g"`.

`22.9"g"xx0.791="actual yield"/color(red)cancel(color(black)(("22.9 g")))^1xxcolor(red)cancel(color(black)(("22.9 g")))^1="18.1 g Cl"_2"`

The actual yield is `"18.1 g Cl"_2"`. (rounded to three significant figures)