Chapter 3, Problem 4KSP

How much of the excess reactant remains when the reaction is complete?

(a)

14.37g

(b)

235.0 g

(c)

78.56 g

(d)

83.96 g

(e)

41.98 g

Interpretation Introduction

Interpretation:

The amount of the reactant ${\text{(Ca}}_{\text{3}}{\text{P}}_{\text{2}})$ that remains after the completion of the reaction is to be calculated.

Concept introduction:

Limiting reagent is the reagent that limits the amount of product during the reaction. Actually, it determines the product, as it is present in a lesser amount than required. Other reagents will remain present in excess.

Excess reactant is the reactant which is present in larger amount than required.

The concept of determining the amount of the reactant remaining at the end of a chemical reaction is based on the observations that are as follows:

1. Calculating the amount of reactant used in the reaction by comparing the ratio of the given reactants.

2. Subtracting the amount of reactant used from the total amount of reactantstaken for the reaction.

3. Converting the moles into mass.

Moles can be calculated as

$\text{Moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

The mass of a compound can be calculated as

$\text{Mass}\text{of compound}\text{= number of moles}\times \text{molar mass}$

The molar mass can be calculated by:

$\text{Molar}\text{mass}\text{=}\text{number}\text{of}\text{element}\text{×}\text{atomic}\text{mass}\text{of}\text{element}$

Example: $\begin{array}{l}{\text{Molar mass of H}}_{\text{2}}\text{O=}\text{number}\text{of}\text{atoms}\text{of}\text{H}\text{×}\text{atomic}\text{mass}\text{of}\text{H}\\ \text{+number}\text{of}\text{atoms}\text{of}\text{O}\text{×}\text{atomic}\text{mass}\text{of}\text{O}\end{array}$

Answer to Problem 4KSP

Solution: Option (a).

Explanation of Solution

Reason for the correct option:

Mass of calcium phosphide, ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}=225.0\text{ g}$. Mass of water, ${\text{H}}_{\text{2}}\text{O = 125}\text{.0 g}$. $$

The balanced equation is as follows:

${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}(s)+\text{6}{H}_2\text{O}(l)\to 3Ca{(OH)}_2(aq)+2P{H}_3(g)$

Then, from the above equation, the molar mass of the reactants ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ and $H_2\text{O}$, and for the product $Ca{(OH)}_2$, is determined by using the atomic weight of calcium, phosphorus, hydrogen and oxygen:

$\begin{array}{c}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}=3(40.08)+\text{2 (30}\text{.97)}\\ =182.18\text{ g/mol}\\ {\text{H}}_{\text{2}}\text{O}=2(1.01)+\text{1 (16}\text{.00)}\\ =18.02\text{ g/mol}\end{array}$

With the help of the above equations, the moles of ${\text{Ca(OH)}}_{\text{2}}$ that can be formed by each reactant is calculated by using the above formula:

$\text{Moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\begin{array}{c}{\text{Mole of Ca}}_{\text{3}}{\text{P}}_{\text{2}}=\frac{\text{225}\text{.0 }\cancel{\text{g}}}{\text{182}\text{.18}\frac{\cancel{\text{g}}}{\text{mol}}}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}\\ \simeq \text{1}{\text{.24 mol Ca}}_{\text{3}}{\text{P}}_{\text{2}}\end{array}$

The moles of ${\text{H}}_2\text{O}$ that can be formed by each reactant is calculated as

$\text{Moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\begin{array}{c}{\text{Mole of H}}_{\text{2}}\text{O}=\frac{\text{125}\text{.0 }\cancel{\text{g}}}{\text{18}\text{.02 }\cancel{\text{g}}\text{/mol}}{\text{H}}_{\text{2}}\text{O}\\ =6.94{\text{ mol H}}_{\text{2}}\text{O}\end{array}$

By comparingboth the reactants

${\text{1 mol Ca}}_{\text{3}}{\text{P}}_{\text{2}}\simeq {\text{6 mol H}}_{\text{2}}\text{O}$

The comparison of the ratio between ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ is as follows:

$\begin{array}{c}\frac{\left({\text{1 mol Ca}}_{\text{3}}{\text{P}}_{\text{2}}\right)}{\left({\text{6 mol H}}_{\text{2}}\text{O}\right)}=\frac{\left(x{\text{ mol Ca}}_3{\text{P}}_2\right)}{\left(6.94{\text{ mol H}}_2\text{O}\right)}\\ x\cancel{{\text{mol Ca}}_3{\text{P}}_2}=\left(6.94\cancel{{\text{mol H}}_2\text{O}}\right)\times \frac{\left(1\cancel{{\text{mol Ca}}_3{\text{P}}_2}\right)}{\left(6\cancel{{\text{mol H}}_2\text{O}}\right)}\\ =1.16\end{array}$

Therefore, the total amount of ${\text{Ca}}_3{\text{P}}_2$ usedis $1.16\text{ mol}$.

Therefore, the remaining reactant iscalculatedas

$\begin{array}{c}{\text{Total Ca}}_{\text{3}}{\text{P}}_{\text{2}}-{\text{Used Ca}}_{\text{3}}{\text{P}}_{\text{2}}=1.24\text{ mol}-\text{1}\text{.16 mol}\\ =\text{0}\text{.08 mol }\end{array}$

Converting moles to mass as follows:

$\begin{array}{c}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}\text{ (in g)}=\left(\text{0}\text{.08 }\cancel{\text{mol}}{\text{ Ca}}_{\text{3}}{\text{P}}_{\text{2}}\right)\times \frac{\text{182}\text{.18 g}}{\cancel{\text{mol}}}\\ \simeq \text{14}{\text{.37 g Ca}}_{\text{3}}{\text{P}}_{\text{2}}\end{array}$

Thus, $\text{14}{\text{.37 g Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ remains after the completion of the reaction.

Hence, option (a) is correct.

Reason for incorrect options:

Option (b) is incorrect because the conversion of Fahrenheit into degree Celsius is correct, but the conversion of Celsius into Kelvin is wrong.

Option (c) is incorrect because, on solving with the help of theabove equations, the answer does not match option (b).

Option (d) is incorrect because, on solving with the help of the above equations, the answer does not match option (d).

Option (e) is incorrect because, on solving with the help of the above equations, the answer does not match option (e).

Hence, options(b), (c), (d), and (e) are incorrect.