SATHEE: Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry - Result Question 40

####40. A mixture of $2.3 g$ formic acid and $4.5 g$ oxalic acid is treated with conc. $H_{2} S O_{4}$ . The evolved gaseous mixture is passed through $K O H$ pellets. Weight (in g) of the remaining product at STP will be

(a) 1.4

(b) 3.0

(d) 2.8

[2018]

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Solution:

(d) $H C O O H \to_{\begin{matrix} Deydrating \\ agent \end{matrix}}^{H_{2} S O_{4}} C O + H_{2} O$

$\begin{matrix} At start \\ (moles) \end{matrix} = \frac{2.3}{46} = \frac{1}{20} \begin{matrix} [H_{2} O absorbed by H_{2} S O . \\ 0 & 0 \end{matrix}$

$\begin{aligned} \begin{array}{c} Final moles & 0 & \frac{1}{20} & \frac{1}{20} \end{array} \\ H_{2} C_{2} O_{4} \overset{H_{2} S O_{4}}{\to} C O + C O_{2} + H_{2} O \\ At start = \frac{4.5}{90} = \frac{1}{20} \\ (moles) \\ [H_{2} O absorbed by H_{2} S O_{4}] \\ \frac{1}{20} \frac{1}{20} \frac{1}{20} \end{aligned}$

$C O_{2}$ is absorbed by $K O H$ .

So, the remaning product is only $C O$ .

Moles of CO formed from both reactions

$= \frac{1}{20} + \frac{1}{20} = \frac{1}{10}$

Left mass of $C O =$ moles $\times$ molar mass

$= \frac{1}{10} \times 28 = 2.8 g$

Some Basic Concepts of Chemistry - Result Question 40

Solution:

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Some Basic Concepts of Chemistry - Result Question 40

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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