Q.

One mole of mixture of CO and CO2 requires exactly 20gm. NaOH to convert all the CO2 into Na2CO3. How many more grams of NaOH would it require for conversion in to Na2CO3, if the mixture is completely oxidised into CO2

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a

40g

b

80g

c

60g

d

20g

answer is B.

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Detailed Solution

Number of moles of CO2 and CO in the mixture = 1

Weight of NaOH required to convert all CO2 into Na2CO3 = 20g

\large \mathop {C{O_2}}\limits_{'{X_1}'mole}^{1mole} + \mathop {2NaOH}\limits_{20g}^{2 \times 40g} \to N{a_2}C{O_3} + {H_2}O\
\large X_1 = \frac{20 \times 1}{2 \times 40} = 0.25mole

 

moles of CO in the initial mixture = (1 - 0.25)= 0.75

Let us assume that all this CO gets Oxidized to CO2 and then reacts with NaOH to get converted to Na2CO3

\large \mathop {CO}\limits_{0.75mole}^{1mole} + \frac{1}{2}{O_2} \to \mathop {C{O_2}}\limits_{'{X_2}'moles}^{1mole}

X2 = 0.75 moles

moles of CO2 formed due to oxiditation of 'CO' = 0.75

\large \mathop {C{O_2}}\limits_{0.75ml}^{1mole} + \mathop {2NaOH}\limits_{'{X_3}'g}^{2 \times 40g} \to N{a_2}C{O_3} + {H_2}O\
\large X_3 = (2 \times 40 \times 0.75)g

= 60g

Wt of NaOH required to neutralize the CO2 obtained by the oxidation of CO = 60g

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