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))#

Question

The reaction is:

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

Ellie Andrews · Accepted Answer

#"0.35 moles Cl"_2#

Examine the balanced chemical equation for this redox reaction first.

#"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.

Your objective at this stage is to determine which of the two reactants functions as a limiting reagent.

To ensure that every mole of manganese dioxide participates in the reaction, you must 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"#

You can say that there will be excess hydrochloric acid because you have more moles than you need.

Because manganese dioxide will be totally consumed in the reaction, it will function as a limiting reagent.

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 conditions that you were given for temperature and pressure are interesting because they match the most recent 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 are aware that the response resulted in

#"7.056 dm"^3 = "7.056 L"#

of chlorine gas at STP, indicating that the reaction's real yield 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#

This can be used to determine the reaction's percent yield.

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

Enter your values to obtain

#"% 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)|)))#

Two sig figs are used to round the answers.

Peyton Adams · Answer

undefined Calculate moles of Cl2 produced: $n = \frac{volume}{molar\space volume} $

Find the limiting reactant: $ \frac{moles\space given}{coefficient\space in\space the\space balanced\space equation} $

Use the limiting reactant to find moles of Cl2: $ moles\space of\space limiting\space reactant \times coefficient\space of\space Cl2 $

Convert moles of Cl2 to grams using molar mass: $ moles\space of\space Cl2 \times molar\space mass\space of\space Cl2 $

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))#

The reaction is:

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