Chapter 3, Problem 3KSP

How much $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ can be produced?

(a)

91.51 g

(b)

274.5 g

(c)

513.8 g

(d)

85.63 g

(e)

257.0 g

Interpretation Introduction

Interpretation:

The amount of ${\text{Ca(OH)}}_{\text{2}}$ produced at the end of the reaction is to be determined.

Concept introduction:

The concept of determining the amount of product formed at the end of a chemical reaction is based on identifying the limiting reactant in that reaction. This is done as follows:

1. Balancing the given chemical equation.

2. Determining the molar mass of each compound.

3. Converting the mass into moles and vice versa by using stoichiometric factors.

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 3KSP

Solution: Option (e).

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:

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

$\begin{array}{c}{\text{Ca(OH)}}_{\text{2}}=\text{1 (40}\text{.08)}+\text{2 (16}\text{.00}+\text{1}\text{.01)}\\ =\text{74}\text{.10 g/mol}\end{array}$

According to the above equation:

$\begin{array}{c}{\text{1 mol Ca}}_{\text{3}}{\text{P}}_{\text{2}}\simeq {\text{3 mol Ca(OH)}}_{\text{2}}\\ {\text{6 mol H}}_{\text{2}}\text{O}\simeq {\text{3 mol Ca(OH)}}_{\text{2}}\end{array}$

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 givenformula:

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

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

$\begin{array}{c}{\text{So, Mole of Ca(OH)}}_{\text{2}}=\left(\text{1}\text{.24 }\cancel{{\text{mol Ca}}_{\text{3}}{\text{P}}_{\text{2}}}\right)\times \frac{\left({\text{3 mol Ca(OH)}}_{\text{2}}\right)}{\left(\text{1}\cancel{{\text{ mol Ca}}_{\text{3}}{\text{P}}_{\text{2}}}\right)}\\ =\text{3}{\text{.72 mol Ca(OH)}}_{\text{2}}\end{array}$

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

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

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

$\begin{array}{c}{\text{So, Mole of Ca(OH)}}_{\text{2}}=\left(\text{6}\text{.94 }\cancel{{\text{mol H}}_{\text{2}}\text{O}}\right)\times \frac{\left({\text{3 mol Ca(OH)}}_{\text{2}}\right)}{\left(\text{6 }\cancel{{\text{mol H}}_{\text{2}}\text{O}}\right)}\\ =\text{3}{\text{.47 mol Ca(OH)}}_{\text{2}}\end{array}$

Now, by comparing the valuesfrom both the reactants, the moles of ${\text{Ca(OH)}}_{\text{2}}$ formed by ${\text{H}}_{\text{2}}\text{O}$ is low, and so, the limiting reactantis ${\text{H}}_{\text{2}}\text{O}$. Therefore, by converting this value of molar mass to mass, the mass of ${\text{Ca(OH)}}_{\text{2}}$ can be calculated as

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

$\begin{array}{c}{\text{Mass of Ca(OH)}}_2\text{= }\left(\text{3}\text{.47 }\cancel{\text{mol}}{\text{ Ca(OH)}}_{\text{2}}\right)\times \frac{\left(\text{74}\text{.1 g}\right)}{\cancel{\text{mol}}}\\ \simeq \text{ 257}{\text{.0 g Ca(OH)}}_2\end{array}$

Therefore, the total amount of ${\text{Ca(OH)}}_{\text{2}}$ obtained from the above reaction is $257.0\text{ g}$.

Hence, option (e) is correct.

Reasons for the incorrect options:

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

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

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

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

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