# Question #761b8

Jun 8, 2016

#### Answer:

The best mixture is ${\text{0.65 g CaC}}_{2}$ and ${\text{46 mL 3% H"_2"O}}_{2}$.

#### Explanation:

WARNING! Long answer.

You want to generate just enough ${\text{C"_2"H}}_{2}$ and ${\text{O}}_{2}$ in a stoichiometric ratio to fill a ${\text{42 in}}^{3}$ cylinder.

I will arbitrarily assume that the pressure is 1 atm and the temperature is 20 °C.

Volume of cylinder:

$\text{Volume of cylinder" = 42 color(red)(cancel(color(black)("in"^3))) × "2.54 cm"/(1 color(red)(cancel(color(black)("in"))))^3 = "688 cm"^3 = "688 mL}$

The balanced equations:

(a) ${\text{CaC"_2 + "2H"_2"O" → "C"_2"H"_2 + "Ca(OH)}}_{2}$

(b) $2 \text{C"_2"H"_2 + "5O"_2 → "4CO"_2 + "2H"_2"O}$

(c) ${\text{2H"_2"O"_2 → "2H"_2"O" + "O}}_{2}$

Volumes of ${\text{O}}_{2}$ and ${\text{C"_2"H}}_{2}$ required:

We see from Equation (b) that we need ${\text{5 mol O}}_{2}$ for every ${\text{2 mol C"_2"H}}_{2}$.

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.

${V}_{\text{C₂H₂" = 2/7 × "688 mL" = "197 mL}}$

${V}_{\text{O₂" = 5/7 × "688 mL" = "492 mL}}$

Mass of miners' grade ${\text{CaC}}_{2}$

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

${\text{Moles of C"_2"H"_2 = (PV)/(RT) = (1 color(red)(cancel(color(black)("atm"))) × 0.197 color(red)(cancel(color(black)("L"))))/("0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 293.15 color(red)(cancel(color(black)("K")))) = "0.008 17 mol C"_2"H}}_{2}$

${\text{Moles of CaC"_2 = "0.008 19" color(red)(cancel(color(black)("mol C"_2"H"_2))) × ("1 mol CaC"_2)/(1 color(red)(cancel(color(black)("mol C"_2"H"_2)))) = "0.008 17 mol CaC}}_{2}$

${\text{Mass of CaC"_2 = "0.008 17" color(red)(cancel(color(black)("mol CaC"_2))) × ("64.10 g CaC"_2)/(1 color(red)(cancel(color(black)("mol CaC_2")))) = "0.524 g CaC}}_{2}$

I assume that miner's grade ($\text{MG}$) ${\text{CaC}}_{2}$ is 80 % pure. Then

${\text{Mass of MG" = 0.524 color(red)(cancel(color(black)("g CaC"_2))) × "100 g MG"/(80 color(red)(cancel(color(black)("g CaC"_2)))) = "0.65 g MG CaC}}_{2}$

Volume of ${\text{H"_2"O}}_{2} :$

${\text{Moles of O"_2 = (PV)/(RT) = (1 color(red)(cancel(color(black)("atm"))) × 0.492 color(red)(cancel(color(black)("L"))))/("0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 293.15 color(red)(cancel(color(black)("K")))) = "0.0204 mol O}}_{2}$

${\text{Moles of H"_2"O"_2 = "0.0204" color(red)(cancel(color(black)("mol O"_2))) × ("2 mol H"_2"O"_2)/(1 color(red)(cancel(color(black)("mol O"_2)))) = "0.0409 mol H"_2"O}}_{2}$

${\text{Mass of H"_2"O"_2 = "0.0409" color(red)(cancel(color(black)("mol H"_2"O"_2))) × ("34.01 g H"_2"O"_2)/(1 color(red)(cancel(color(black)("mol H"_2"O_2")))) = "1.39 g H"_2"O}}_{2}$

$\text{Mass of H"_2"O"_2color(white)(l) "solution" = 1.39 color(red)(cancel(color(black)("g H"_2"O"_2))) × "100 g solution"/(3 color(red)(cancel(color(black)("g H"_2"O"_2)))) = "46.3 g solution}$

$\text{Volume of solution" = 46.25 color(red)(cancel(color(black)("g solution"))) × "1 mL solution"/(1 color(red)(cancel(color(black)("g solution")))) = "46 mL solution}$