# Question f25c6

Mar 6, 2017

The mixture contains ${\text{16 cm}}^{3}$ of methane and ${\text{20 cm}}^{3}$ of acetylene.

#### Explanation:

We can use Gay-Lussac's Law of Combining Volumes to solve this problem.

The equations are

$\textcolor{w h i t e}{l} \text{CH"_4 + "2O"_2 → "CO"_2 + "2H"_2"O}$
${\text{1 cm"^3color(white)(m)"2 cm"^3color(white)(m)"1 cm}}^{3}$

$\text{2C"_2"H"_2 + "5CO"_2 → "4CO"_2 + "2H"_2"O}$
$\textcolor{w h i t e}{l} {\text{2 cm"^3color(white)(mll)"5 cm"^3color(white)(mll)"4 cm}}^{3}$

${V}_{\textrm{m e t h a \ne}} + {V}_{\textrm{a c e t y \le \ne}} = 36 \textcolor{w h i t e}{l} {\text{cm}}^{3}$

Let the volume of methane be $x \textcolor{w h i t e}{l} {\text{cm}}^{3}$.

Then the volume of acetylene is $\left(36 - x\right) \textcolor{w h i t e}{l} {\text{cm}}^{3}$.

V_text(CO₂)color(white)(l) "from methane" + V_text(CO₂) color(white)(l)"from acetylene" = V_text(CO₂) color(white)(l)"total"#

$x \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{cm"^3color(white)(l) "CH"_4))) × (1 color(red)(cancel(color(black)("cm"^3color(white)(l) "CO"_2))))/(1 color(red)(cancel(color(black)("cm"^3 color(white)(l)"CH"_4)))) + (36 - x) color(red)(cancel(color(black)("cm"^3 color(white)(l)"C"_2"H"_2))) × (4 color(red)(cancel(color(black)("cm"^3 color(white)(l)"CO"_2))))/(2 color(red)(cancel(color(black)("cm"^3color(white)(l) "C"_2"H"_2)))) = 56 color(red)(cancel(color(black)("cm"^3 color(white)(l)"CO}}_{2}}}}$

$x + 2 \left(36 - x\right) = 56$

$x + 72 - 2 x = 56$

$x = 16$

${\text{Volume of methane" = "16 cm}}^{3}$ and

${\text{Volume of acetylene" = "(36 - 16) cm"^3 = "20 cm}}^{3}$