Chapter 13, Problem 140AP

he following demonstration takes place in a two-step process:

rst, solid calcium carbide $\left({\text{CaC}}_{\text{2}}\text{j}\right)$reacts with liquid water to produce acetylene gas $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\right)$and aqueous calcium hydroxide. Second the acetylene gas produced is then ignited with a match, causing the combustion reaction of acetylene with oxygen gas to produce gaseous carbon dioxide and gaseous water.

Write the balanced equations for each reaction that is occurring, including all phases.

If a $\text{1}00.0-\text{g}$sample of calcium carbide $\left({\text{CaC}}_{\text{2}}\right)$is initially reacted with $\text{5}0.0\text{ g}$of water, which reactant is limiting?

Now imagine that the final gases produced are collected in a large bulkier and allowed to cool to room temperature. Using the information from part b ( $\text{l}00.0\text{ g}$of ${\text{Cec}}_{\text{2}}$reacting with $\text{5}0.0\text{ g}$of ${\text{H}}_{\text{2}}\text{O}$), how many liters of carbon dioxide gas were produced in the balloon at a pressure of $\text{1}.00$atm and $\text{25 }°\text{C}$?

Interpretation Introduction

(a)

Interpretation:

The balance chemical equation for the combustion of ethyne and reaction of calcium carbide to produce ethyne.

Concept Introduction:

When the number of atoms of an element on the product side and on the reactant side is equal then such reaction is said to be a balanced chemical equation.

Answer to Problem 140AP

The balanced chemical reaction is given as:

${\text{ CaC}}_{\text{2(s)}}{\text{+ 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{ C}}_{\text{2}}{\text{H}}_{\text{2(g)}}\text{+ Ca}{\left(\text{OH}\right)}_{\text{2(aq)}}$

${\text{2C}}_{\text{2}}{\text{H}}_{\text{2(g)}}{\text{+5O}}_{\text{2(g)}}\to {\text{4CO}}_{\text{2(g)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\text{(g)}}$.

Explanation of Solution

The balance equation indicates that all the number of same atoms must be equal at both sides. Hence we have to write the reactants and products of the reaction and balance each atom at both sides.

${\text{ CaC}}_{\text{2(s)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{ C}}_{\text{2}}{\text{H}}_{\text{2(g)}}\text{+ Ca}{\left(\text{OH}\right)}_{\text{2(aq)}}$

In this reaction equation, number of C and Ca is balanced at both sides. Hence we have to balance H and O. To balance H; add 2 as coefficient with H₂ O that changes the equation as;

${\text{ CaC}}_{\text{2(s)}}{\text{+ 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{ C}}_{\text{2}}{\text{H}}_{\text{2(g)}}\text{+ Ca}{\left(\text{OH}\right)}_{\text{2(aq)}}$

Now O and H both are balanced hence it is a balance equation.

Similarly in second reaction equation;

${\text{ C}}_{\text{2}}{\text{H}}_{\text{2(g)}}{\text{+ O}}_{\text{2(g)}}\to {\text{ CO}}_{\text{2(g)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

Let’s balance Carbon at both sides;

${\text{2C}}_{\text{2}}{\text{H}}_{\text{2(g)}}{\text{+ O}}_{\text{2(g)}}\to {\text{4CO}}_{\text{2(g)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

Now we have to balance H and O at both sides. To balance O, add 5 as coefficient with O₂ and 2 as coefficient with H₂ O.

${\text{2C}}_{\text{2}}{\text{H}}_{\text{2(g)}}{\text{+5O}}_{\text{2(g)}}\to {\text{4CO}}_{\text{2(g)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\text{(g)}}$.

Interpretation Introduction

(b)

Interpretation:

Identify the limiting reactant if 100.0- g sample of calcium carbide is initially reacted with 50.0 g of water.

Concept Introduction:

The reactant that forms less amount of product is called as limiting reagent and which limits the reaction and determine the amount of product formed.

Answer to Problem 140AP

Water forms less amount of ethyne so water is limiting reagent.

Explanation of Solution

${\text{ CaC}}_{\text{2(s)}}{\text{+ 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{ C}}_{\text{2}}{\text{H}}_{\text{2(g)}}\text{+ Ca}{\left(\text{OH}\right)}_{\text{2(aq)}}$

Molar mass of CaC₂ = 64.09 g/mol

Molar mass of H₂ O = 18.0 g/mol

Mass of CaC₂ = 100 g

Mass of H₂ O = 50.0 g

Number of moles of CaC₂ = $\frac{\text{100 g}}{\text{64}\text{.09 g/mol}}\text{=1}\text{.56 mol}$

Number of moles of water = $\frac{\text{50}\text{.0 g}}{\text{18}\text{.0 g/mol}}\text{=2}\text{.78 mol}$

According to balance chemical equation, 1 mole of ethyne is formed by 2 moles of H₂ O and 1 mole of ethyne is formed by 1 mole of calcium carbide.

$\frac{\text{1mole of ethyne}}{1{\text{ mole of CaC}}_2}\times \text{ 1}{\text{.56 mole of CaC}}_2\text{=1}\text{.56 mole of ethyne}$

$\frac{\text{1mole of ethyne}}{{\text{2 mole of H}}_2\text{O}}\times \text{ 2}{\text{.78 mole of H}}_2\text{O =1}\text{.39 mole of ethyne}$

Hence, H₂ O forms less amount of ethyne so water is limiting reagent.

Interpretation Introduction

(c)

Interpretation:

The volume of carbon dioxide gas which is produced in the balloon at a pressure of 1.00 atm and 25° C.

Concept Introduction:

The ideal gas equation is given by:

PV = nRT

Where, P = pressure

V = volume

n = number of moles

T = temperature

R = gas constant

By knowing the value of pressure, temperature and number of moles, one can identify the value of volume.

Answer to Problem 140AP

Volume of carbon dioxide produced is 68.01 L.

Explanation of Solution

Given information:

• P = 1 atm.
• T = 25 °C = 25+373 = 298 K.
• n = 2.78 moles (from part b).
• R = 0.0821 L.atm / k.mol.

The given reaction is

${\text{CaC}}_{\text{2(s)}}{\text{+ 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{ C}}_{\text{2}}{\text{H}}_{\text{2(g)}}\text{+ Ca}{\left(\text{OH}\right)}_{\text{2(aq)}}$

Put the given values in ideal gas equation.

PV = nRT

$\begin{array}{l}\text{1 atm }\times \text{ V = 2}\text{.78 moles }\times \text{ 0}\text{.0821 L }\text{. atm / K }\text{.mol }\times \text{ 298 K}\\ \text{V=}\frac{\text{2}\text{.78 moles }\times \text{ 0}\text{.0821 L }\text{. atm / K }\text{.mol }\times \text{ 298 K}}{\text{1 atm}}=68.01\text{ L}\end{array}$.