Chapter 3, Problem 1KSP

Calculate the mass of water produced in the example.

(a)

21.02 g

(b)

10.51 g

(c)

11.61 g

(d)

11. 75 g

(e)

5400 g

Use the following information to answer questions 3.2, 3.3, and 3.4.

Calcium phosphide $\left({\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}\right)$ and water react to form calcium hydroxide and phosphine $\left({\text{PH}}_{\text{3}}\right)$. In a particular experiment. 225.0 g ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ and 125.0 g water are combined.

${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}\left(s\right){\text{ + H}}_{\text{2}}\text{O}\left(l\right)\underset{}{\to }\text{ Ca}{\left(\text{OH}\right)}_{\text{2}}\left(aq\right){\text{ + PH}}_{\text{3}}\left(g\right)$

(Don't forget to balance the equation.)

Interpretation Introduction

Interpretation:

The mass of water produced in the given example 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 be present in excess.

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

The required amount of ${\text{N}}_2{\text{O}}_4$ can be calculated by:

$\text{Mass}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}=\frac{\text{number}\text{of}\text{moles}\times \text{molar}\text{mass}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}{\text{number}\text{of}\text{moles}\times \text{molar}\text{mass}\text{of}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}}\text{×}\text{mass}\text{of}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$

The required amount of ${\text{H}}_2\text{O}$ can be calculated by:

$\text{Mass}\text{of}{\text{H}}_{\text{2}}\text{O}=\frac{\text{number}\text{of}\text{moles}\times \text{molar}\text{mass}\text{of}{\text{H}}_{\text{2}}\text{O}}{\text{number}\text{of}\text{moles}\times \text{molar}\text{mass}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{×}\text{mass}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$

Answer to Problem 1KSP

Solution: Option (b).

Explanation of Solution

Reason for thecorrect option:

See the balanced chemical equation:

${\text{N}}_2{\text{H}}_4{\text{+2N}}_2{\text{O}}_4\to 6{\text{NO +2H}}_2\text{O}$

Here, ${\text{ 32 g N}}_2{\text{H}}_4$ reacts with $184{\text{ g N}}_2{\text{O}}_4$ to produce $180\text{ g NO}$ and $36{\text{ g H}}_2\text{O}$.

For $32{\text{ g N}}_2{\text{H}}_4$, $184{\text{ g N}}_2{\text{O}}_4$ is required, so for $10.45{\text{ g N}}_2{\text{H}}_4$, the required amount of ${\text{N}}_2{\text{O}}_4$ is

$\text{Mass}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}=\frac{\text{number}\text{of}\text{moles}\times \text{molar}\text{mass}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}{\text{number}\text{of}\text{moles}\times \text{molar}\text{mass}\text{of}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}}\text{×}\text{mass}\text{of}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$

$\begin{array}{l}\text{32}\text{g}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{reacts}\text{with}\text{184}\text{g}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\\ \text{1}\text{g}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\to \frac{\text{184}\text{g}}{\text{32}\text{g}}\\ \text{10}\text{.}\text{45}\text{g}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\to \frac{\text{184}\text{g}}{\text{32}\text{g}}\text{×10}\text{.}\text{45}\text{g}\end{array}$

$\begin{array}{c}\frac{(10.45)\times (184)}{(32)}=\frac{1922.8}{32}\\ =60.08\text{ g}\end{array}$

But the available amount of ${\text{N}}_2{\text{O}}_4$ is $53.68\text{ g}$, therefore it is the limiting reagent. And it will determine the number of products.

Since $184{\text{ g N}}_2{\text{O}}_4$ produces $36{\text{ g H}}_2\text{O}$, $53.68\text{ g}$ ${\text{N}}_2{\text{O}}_4$ will produce ${\text{H}}_2\text{O}$:

$\text{Mass}\text{of}{\text{H}}_{\text{2}}\text{O}=\frac{\text{number}\text{of}\text{moles}\times \text{molar}\text{mass}\text{of}{\text{H}}_{\text{2}}\text{O}}{\text{number}\text{of}\text{moles}\times \text{molar}\text{mass}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{×}\text{mass}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$

$\begin{array}{l}\text{184}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{produces}\text{36}\text{g}{\text{H}}_{\text{2}}\text{O}\\ \text{1}\text{g}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\to \frac{\text{36}\text{g}}{\text{184}\text{g}}\\ 53.68\text{g}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\to \frac{\text{36}\text{g}}{\text{184}\text{g}}\times 53.68\text{g}\end{array}$

$\begin{array}{c}\frac{(53.68)\times (36)}{184}=\frac{1932.48}{184}\\ =10.50{\text{ g H}}_2\text{O}\end{array}$

Hence, option (b) is correct.

Reasons for the incorrect options:

Option (a) is incorrect because according to the above relation, the value $21.02\text{ g}$ is not obtained.

Option (c) is incorrect because according to the above relation, the value $11.61\text{ g}$ is not obtained.

Option (d) is incorrect because according to the above relation, the value $11.75\text{ g}$ is not obtained.

Option (e) is incorrect because according to the above relation, the value $5.400\text{ g}$ is not obtained.

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