Question #d3347

1 Answer
Sep 1, 2016

Answer:

Here's what I got.

Explanation:

Judging from the amounts of methane and chlorine gas given by the problem, you can say that this reaction will produce chloromethane, #"CH"_3"Cl"#, and hydrogen chloride, #"HCl"#

#"CH"_ (4(g)) + "Cl"_ (2(g)) stackrel(color(white)(acolor(blue)(nu)aaa))(->) "CH"_ 3"Cl"_ ((g)) + "HCl"_ ((g))#

Notice that the reaction consumes methane and chlorine gas in a #1:1# mole ratio. At this point, you should convert the masses given to you to moles to see which reactant, if any, acts as a limiting reagent.

To do that, use the molar masses of methane and of chlorine gas

#23.5 color(red)(cancel(color(black)("g"))) * "1 mole CH"_4/(16.04color(red)(cancel(color(black)("g")))) = "1.465 moles CH"_4#

#64 color(red)(cancel(color(black)("g"))) * "1 mole Cl"_2/(70.906color(red)(cancel(color(black)("g")))) = "0.9026 moles Cl"_2#

Now, in order for the reaction to consume all the moles of methane, you need to provide #1.465# moles of chlorine gas, as given by the aforementioned #1:1# mole ratio.

Since you have fewer moles of chlorine gas available, chlorine gas will act as a limiting reagent, i.e. it will be completely consumed before all the moles of methane get the chance to react.

In other words, you have an excess of methane. The reaction will consume all the moles of chlorine gas and

#0.9026 color(red)(cancel(color(black)("moles Cl"_2))) * "1 mole CH"_4/(1color(red)(cancel(color(black)("mole Cl"_2)))) = "0.9026 moles CH"_4#

You can thus say that you have an excess of

#overbrace("1.465 moles CH"_4)^(color(darkgreen)("what you start with")) - overbrace("0.9026 moles CH"_4)^(color(purple)("what gets consumed")) = "0.5624 moles CH"_4#

Convert this to grams to find the mass of methane that does not take part in the reaction

#0.5624 color(red)(cancel(color(black)("moles CH"_4))) * "16.04 g"/(1color(red)(cancel(color(black)("mole CH"_4)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("9.0 g")color(white)(a/a)|)))#

Now, notice that the reaction produces chloromethane and hydrogen chloride in #1:1# mole ratios with the two reactants.

This means that in theory, for every mole of methane and of chlorine gas that take part in the reaction, you get #1# mole of chloromethane and #1# mole of hydrogen chloride.

In other words, a reaction that has a #100%# yield will produce chloromethane and hydrogen chloride in #1:1# mole ratios.

You can thus say that the reaction will theoretically produce

#0.9026 color(red)(cancel(color(black)("moles CH"_ 4))) * ("1 mole CH"_ 3"Cl")/(1color(red)(cancel(color(black)("mole CH"_ 4)))) = "0.9026 moles CH"_ 3"Cl"#

#0.9026 color(red)(cancel(color(black)("moles CH"_4))) * "1 mole HCl"/(1color(red)(cancel(color(black)("mole CH"_4)))) = "0.9026 moles HCl"#

However, you are told that the reaction has a #73%# yield. This tells you that for every #100# moles of a product that can be theoretically produced by the reaction, you only get #73# moles.

This means that the reaction will actually produce

#0.9026 color(red)(cancel(color(black)("moles CH"_3"Cl"))) * overbrace(("73 moles CH"_ 3"Cl")/(100color(red)(cancel(color(black)("moles CH"_3"Cl")))))^(color(brown)("73% yield")) = "0.6589 moles CH"_3"Cl"#

#0.9026 color(red)(cancel(color(black)("moles HCl"))) * overbrace(("73 moles HCl")/(100color(red)(cancel(color(black)("moles HCl")))))^(color(brown)("73% yield")) = "0.6589 moles HCl"#

Once again, use the molar masses of the two products to convert the moles to grams

#0.6589 color(red)(cancel(color(black)("moles CH"_3"Cl"))) * "119.38 g"/(1color(red)(cancel(color(black)("mole CH"_3"Cl")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("79 g CH"_3"Cl")color(white)(a/a)|)))#

#0.6589 color(red)(cancel(color(black)("moles HCl"))) * "36.46 g"/(1color(red)(cancel(color(black)("mole HCl")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("24 g HCl")color(white)(a/a)|)))#

All the values are rounded to two sig figs.