WARNING! Long answer.
You want to generate just enough #"C"_2"H"_2# and #"O"_2# in a stoichiometric ratio to fill a #"42 in"^3# cylinder.
I will arbitrarily assume that the pressure is 1 atm and the temperature is 20 °C.
Volume of cylinder:
#"Volume of cylinder" = 42 color(red)(cancel(color(black)("in"^3))) × "2.54 cm"/(1 color(red)(cancel(color(black)("in"))))^3 = "688 cm"^3 = "688 mL"#
The balanced equations:
(a) #"CaC"_2 + "2H"_2"O" → "C"_2"H"_2 + "Ca(OH)"_2#
(b) #2"C"_2"H"_2 + "5O"_2 → "4CO"_2 + "2H"_2"O"#
(c) #"2H"_2"O"_2 → "2H"_2"O" + "O"_2#
Volumes of #"O"_2# and #"C"_2"H"_2# required:
We see from Equation (b) that we need #"5 mol O"_2# for every #"2 mol C"_2"H"_2#.
Since the ratio of the moles is the same as the ratio of the volumes, 5/7 of the volume is oxygen and 2/7 of the volume is acetylene.
#V_"C₂H₂" = 2/7 × "688 mL" = "197 mL"#
#V_"O₂" = 5/7 × "688 mL" = "492 mL"#
Mass of miners' grade #"CaC"_2#
#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#
#"Moles of C"_2"H"_2 = (PV)/(RT) = (1 color(red)(cancel(color(black)("atm"))) × 0.197 color(red)(cancel(color(black)("L"))))/("0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 293.15 color(red)(cancel(color(black)("K")))) = "0.008 17 mol C"_2"H"_2#
#"Moles of CaC"_2 = "0.008 19" color(red)(cancel(color(black)("mol C"_2"H"_2))) × ("1 mol CaC"_2)/(1 color(red)(cancel(color(black)("mol C"_2"H"_2)))) = "0.008 17 mol CaC"_2#
#"Mass of CaC"_2 = "0.008 17" color(red)(cancel(color(black)("mol CaC"_2))) × ("64.10 g CaC"_2)/(1 color(red)(cancel(color(black)("mol CaC_2")))) = "0.524 g CaC"_2#
I assume that miner's grade (#"MG"#) #"CaC"_2# is 80 % pure. Then
#"Mass of MG" = 0.524 color(red)(cancel(color(black)("g CaC"_2))) × "100 g MG"/(80 color(red)(cancel(color(black)("g CaC"_2)))) = "0.65 g MG CaC"_2#
Volume of #"H"_2"O"_2:#
#"Moles of O"_2 = (PV)/(RT) = (1 color(red)(cancel(color(black)("atm"))) × 0.492 color(red)(cancel(color(black)("L"))))/("0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 293.15 color(red)(cancel(color(black)("K")))) = "0.0204 mol O"_2#
#"Moles of H"_2"O"_2 = "0.0204" color(red)(cancel(color(black)("mol O"_2))) × ("2 mol H"_2"O"_2)/(1 color(red)(cancel(color(black)("mol O"_2)))) = "0.0409 mol H"_2"O"_2#
#"Mass of H"_2"O"_2 = "0.0409" color(red)(cancel(color(black)("mol H"_2"O"_2))) × ("34.01 g H"_2"O"_2)/(1 color(red)(cancel(color(black)("mol H"_2"O_2")))) = "1.39 g H"_2"O"_2#
#"Mass of H"_2"O"_2color(white)(l) "solution" = 1.39 color(red)(cancel(color(black)("g H"_2"O"_2))) × "100 g solution"/(3 color(red)(cancel(color(black)("g H"_2"O"_2)))) = "46.3 g solution"#
#"Volume of solution" = 46.25 color(red)(cancel(color(black)("g solution"))) × "1 mL solution"/(1 color(red)(cancel(color(black)("g solution")))) = "46 mL solution"#
