Question

At 273.15 K and 100kPa, 58.34 g of `"HCl"` reacts with 0.35 mol of `"MnO"_2` to produce `"7.056 dm"^3` of chlorine gas. Calculate the theoretical yield of chlorine. How do I solve this?

The reaction is:

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

Answer 1

`"0.35 moles Cl"_2`

Explanation:

Start by taking a look at the balanced chemical equation for this redox reaction

`"MnO"_ (2(s)) + color(red)(4)"HCl"_ ((aq)) -> "MnCl"_ (2(aq)) + "Cl"_ (2(g)) uarr + 2"H"_ 2"O"_((l))`

Your starting point here is the `1:color(red)(2)` mole ratio that exists between manganese dioxide and hydrochloric acid.

This mole ratio tells you that the reaction will always consume `color(red)(4)` moles of hydrochloric acid for every mole of manganese dioxide that takes part in the reaction.

At this point, your goal is to figure out if one of the two reactants acts as a limiting reagent.

In order for all the moles of manganese dioxide to actually take part in the reaction, you need to have

`0.35color(red)(cancel(color(black)("moles MnO"_2))) * (color(red)(4)color(white)(a)"moles HCl")/(1color(red)(cancel(color(black)("mole MnO"_2)))) = "1.40 moles HCl"`

Use the molar mass of hydrochloric acid to determine how many moles you get in that `"58.4-g"` sample

`58.34 color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.46color(red)(cancel(color(black)("g")))) = "1.6001 moles HCl"`

Since you have more moles of hydrochloric acid than you need, you can say that the acid will be in excess.

Manganese dioxide will act as a limiting reagent, i.e. it will be completely consumed in the reaction.

Now, notice that you have a `1:1` mole ratio between manganese dioxide and chlorine gas. This means that the reaction can theoretically produce one mole of chlorine gas for every mole of manganese dioxide that reacts.

The theoretical yield of the reaction will correspond to a `100%` yield, so you can say that the reaction can theoretically produce

`0.35color(red)(cancel(color(black)("moles MnO"_2))) * "1 mole Cl"_2/(1color(red)(cancel(color(black)("mole MnO"_2)))) = color(green)(|bar(ul(color(white)(a/a)"0.35 moles Cl"_2color(white)(a/a)|)))`

The cool thing about the conditions for pressure and temperature given to you is that they correspond to the current definition of STP.

Under these conditions for pressure and temperature, one mole of any ideal gas occupies `"22.7 L " ->` this is known as the molar volume of a gas at STP.

You know that the reaction produced

`"7.056 dm"^3 = "7.056 L"`

of chlorine gas at STP, which means that the actual yield of the reaction was

`7.056color(red)(cancel(color(black)("L"))) * overbrace("1 mole Cl"_2/(22.7color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas at STP")) = "0.3108 moles Cl"_2`

You can use this to calculate the percent yield of the reaction

`color(blue)(|bar(ul(color(white)(a/a)"% yield" = "what you actually get"/"what you should theoretically get" xx 100 color(white)(a/a)|)))`

Plug in your values to get

`"% yield" = (0.3108 color(red)(cancel(color(black)("moles"))))/(0.35color(red)(cancel(color(black)("moles")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"89%"color(white)(a/a)|)))`

The answers are rounded to two sig figs.