SATHEE: Some Basic Concepts Of Chemistry Question 66

Some Basic Concepts Of Chemistry Question 66

Question: 1 mole of mixture of CO and $C O_{2}$ requires exactly $28 g K O H$ in solution for complete conversion of all the $C O_{2}$ into $K_{2} C O_{3} .$ How much amount more of KOH will be required for conversion into $K_{2} C O_{3} .$ If one mole of mixture is completely oxidized to $C O_{2}$

Options:

A) 112g

B) 84g

C) 56g

D) 28g

Show Answer

Answer:

Correct Answer: B

Solution:

[b] $C O + \frac{1}{2} O_{2} \to C O_{2};$ $C O_{2} + 2 K O H \to K_{2} C O_{3} + H_{2} O$ Moles of $K O H = \frac{28}{56} = 0.50$ It corresponds to 0.25 mol of $C O_{2}$ Hence mol of $C O = 1 - 0.25 = 0.75$ $\equiv$ mole of $C O_{2}$ formed Mol of KOH requred $= 2 \times 0.75 = 1.5$ $= 1.5 \times 56 = 84 g$

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Some Basic Concepts Of Chemistry Question 66

Question: 1 mole of mixture of CO and CO2 requires exactly 28gKOH in solution for complete conversion of all the CO2 into K2CO3. How much amount more of KOH will be required for conversion into K2CO3. If one mole of mixture is completely oxidized to CO2

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Question: 1 mole of mixture of CO and $C O_{2}$ requires exactly $28 g K O H$ in solution for complete conversion of all the $C O_{2}$ into $K_{2} C O_{3} .$ How much amount more of KOH will be required for conversion into $K_{2} C O_{3} .$ If one mole of mixture is completely oxidized to $C O_{2}$