Question

Before small batteries were available, carbide lamps were used for bicycle lights. Acetylene gas, C₂H₂, and solid calcium hydroxide were formed by the reaction of calcium carbide, CaC₂, with water. The ignition of the acetylene gas provided the light. Currently, the same lamps are used by some cavers, and calcium carbide is used to produce acetylene for carbide cannons.

(a) Outline the steps necessary to answer the following question: What volume of C₂H₂at atm and 12.2̊C is formed by the reaction of 15.48g of CaC₂with water?

(b) Answer the question.

Short answer

1. Outlined the necessary steps.

Step 1: Write the balanced equation for the reaction.

Step 2: Determine the number of moles of calcium carbide \[\left( {{\rm{Ca}}{{\rm{C}}_2}} \right)\] used that determine the number of moles of \[{{\rm{C}}_2}{{\rm{H}}_2}\] produced.

Step 3: Find the volume of \[{{\rm{C}}_2}{{\rm{H}}_2}\], using ideal gas equation.

(b) 5.61 L \[{{\rm{C}}_2}{{\rm{H}}_2}\] is produced.

Step-by-step solution

Definition of mass

Mass is a unit of measurement for the amount of matter in a body.

Step 2: Explanation for Part (a)

(a)

Step 1: First, we write the balanced equation for the reaction.

Step 2: Determine the number of moles of calcium carbide \[\left( {{\rm{Ca}}{{\rm{C}}_2}} \right)\] in \[15.48\;{\rm{g}}\] and from that determine the number of moles of \[{{\rm{C}}_2}{{\rm{H}}_2}\] produced.

Step 3: Find the volume of \[{{\rm{C}}_2}{{\rm{H}}_2}\], using ideal gas equation.

Use ideal gas law to get the final answer

(b)

Consider the given information.

\[{\rm{Ca}}{{\rm{C}}_2} + 2{{\rm{H}}_2}{\rm{O}} \to {{\rm{C}}_2}{{\rm{H}}_2} + {\rm{Ca}}{({\rm{OH}})_2}\]

Mass of\[{\rm{Ca}}{{\rm{C}}_2} = 15.48\;{\rm{g}}\]

Molar mass of\[{\rm{Ca}}{{\rm{C}}_2} = 64.1\;{\rm{g}}/{\rm{mol}}\]

Moles of\[{\rm{Ca}}{{\rm{C}}_2}\]:

\begin{aligned}_{{{\rm{C}}_2}{{\rm{H}}_2}}=0.241\;{\rm{molCa}}{{\rm{C}}_2}\times\frac{{1\;{\rm{mol}}{{\rm{C}}_2}{{\rm{H}}_2}}}{{1\;{\rm{molCa}}{{\rm{C}}_{\rm{2}}}}}\\= 0.241\;{\rm{mol}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}T\\={12.2R^\circ}{\rm{C}}+273\\=285.2\;{\rm{K}},P\\=1.005\;{\rm{atm}}\end{aligned}

From the equation, we can see that\[1\]mol of\[{\rm{Ca}}{{\rm{C}}_2}\]produces\[1\]mol of\[{{\rm{C}}_2}{{\rm{H}}_2}\]. So using mole ratio we can find the moles of\[{{\rm{C}}_2}{{\rm{H}}_2}\]produced by\[0.241\;{\rm{molCa}}{{\rm{C}}_2}\]

\begin{aligned}_{{{\rm{C}}_2}{{\rm{H}}_2}}=0.241\;{\rm{molCa}}{{\rm{C}}_2}\times\frac{{1\;{\rm{mol}}{{\rm{C}}_2}{{\rm{H}}_2}}}{{1\;{\rm{molCa}}{{\rm{C}}_{\rm{2}}}}}\\=0.241\;{\rm{mol}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}T\\={12.2R^\circ}{\rm{C}}+273\\=285.2\;{\rm{K}},P\\=1.005\;{\rm{atm}}\end{aligned}

Therefore, finally we can use the ideal gas law:

\begin{aligned}{V_{{{\rm{C}}_2}{{\rm{H}}_2}}}={\left({\frac{{nRT}}{P}}\right)_{{{\rm{C}}_2}{{\rm{H}}_2}}}\\=\frac{{0.241\;{\rm{mol}}\times0.08206\;{\rm{L}}\;{\rm{atmmo}}{{\rm{l}}^{-1}}\;{{\rm{K}}^{-1}}\times285.2\;{\rm{K}}}}{{1.005\;{\rm{atm}}}}\\=5.61\;{\rm{L}}\end{aligned}