Question

What is the change in volume for the reaction of `"16 g CH"_4` and `"16 g O"_2` at `700^@ "C"` and `"1 bar"`? The reaction is `"CH"_4(g) + 1/2"O"_2(g) -> "CO"(g) + 2"H"_2(g)`.

Answer 1

About `"120 L"`.

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For this, since we are asked for the change in volume, it is necessary to solve for the molar volume of an ideal gas at `700^@ "C"` (since it is evidently not `0^@ "C"` like it would be for STP).

`PV = nRT`

`=> V/n = (RT)/P`

`= (("0.083145 L"cdot"bar/mol"cdot"K")("700+273.15 K"))/("1 bar")`

`=` `"80.91 L/mol"`

Given `"16 g"` of both methane and diatomic oxygen, we can then find their `"mol"`s.

`"16 g CH"_4 xx ("1 mol CH"_4)/(12.011 + 4 xx 1.0079 "g CH"_4) = "0.9973 mols CH"_4`

`"16 g O"_2 xx ("1 mol O"_2)/(15.999 xx 2 "g O"_2) = "0.5000 mols O"_2`

No, it is not necessary to halve anything. You have from the masses the same mol/mol ratio as required in the reaction itself. You needed `"1 mol CH"_4(g)` for every `"0.5 mol O"_2(g)`, and that's what you have, basically.

You do have a bit extra `"O"_2` than you need, though, so let's just be particular about it... The limiting reactant is `"CH"_4`, since there is slightly less than twice the mols of `"CH"_4` as `"O"_2`, when the reaction requires exactly twice.

So:

`"0.9973 mols CH"_4 harr "0.4987 mols O"_2`

and `"0.0130 mols O"_2` is in excess. That will remain, but we have to account for that in calculating the mols of the products by using `"CH"_4` instead of `"O"_2` to calculate the mols of products.

By the mol/mol ratio given in the reaction:

`"mols CH"_4 harr "0.9973 mols CO"(g)`

`"mols CH"_4 xx 2 harr "1.9947 mols H"_2(g)`

So, really the next thing we can calculate are the final and initial mols:

`n_(CO) + n_(H_2) = n_2`

`n_(CH_4) + n_(O_2) = n_1`

`=> "0.9973 mols CO" + "1.9947 mols H"_2 = n_2 = "2.9920 mols gas"`

`=> "0.9973 mols CH"_4 + "(0.4987 + 0.0130) mols O"_2 = n_1 = "1.5090 mols gas"`

That means the ratio of the mols can be related for a big picture using Gay-Lussac's Law:

`V_1/(n_1) = V_2/(n_2)`

`V_1/("1.5090 mols") = V_2/("2.9920 mols")`

By the molar volume, we can find each actual volume for the ideal gas combinations.

`V_1/"1.5090 mols" = "80.91 L"/"mol"`

`=> V_1 = "122.093 L"`

`V_2/"2.9920 mols" = "80.91 L"/"mol"`

`=> V_2 = "242.083 L"`

So, the change in volume is then:

`bb(DeltaV = V_2 - V_1)`

`= "242.083 L" - "122.093 L"`

`=` `bb"119.99 L"`

To two sig figs, `color(blue)(DeltaV = "120 L")`.

As a check, we can see whether the ideal gas law is still satisfied.

`PDeltaV stackrel(?" ")(=)DeltanRT`

`("1 bar")("120 L") stackrel(?" ")(=) ("2.9920 - 1.5090 mols gas")("0.083145 L"cdot"bar/mol"cdot"K")("973.15 K")`

`"120 L"cdot"bar" ~~ "119.993 L"cdot"bar"`

Yep, we're good.