Question #f25c6

1 Answer
Mar 6, 2017

Answer:

The mixture contains #"16 cm"^3# of methane and #"20 cm"^3# of acetylene.

Explanation:

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

The equations are

#color(white)(l)"CH"_4 + "2O"_2 → "CO"_2 + "2H"_2"O"#
#"1 cm"^3color(white)(m)"2 cm"^3color(white)(m)"1 cm"^3#

#"2C"_2"H"_2 + "5CO"_2 → "4CO"_2 + "2H"_2"O"#
#color(white)(l)"2 cm"^3color(white)(mll)"5 cm"^3color(white)(mll)"4 cm"^3#

#V_text(methane) + V_text(acetylene) = 36color(white)(l) "cm"^3#

Let the volume of methane be #xcolor(white)(l) "cm"^3#.

Then the volume of acetylene is #(36 - x)color(white)(l) "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 color(red)(cancel(color(black)("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(36-x) = 56#

#x + 72 -2x = 56#

#x = 16#

#"Volume of methane" = "16 cm"^3# and

#"Volume of acetylene" = "(36 - 16) cm"^3 = "20 cm"^3#