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Question: What mass of \[HN{O_3}\] is needed to convert \[5g\] of iodine into iodic acid according to the reac...

What mass of HNO3HN{O_3} is needed to convert 5g5g of iodine into iodic acid according to the reaction:
I2+HNO3→HIO3+NO2+H2O{I_2} + HN{O_3} \to HI{O_3} + N{O_2} + {H_2}O
A. 12.4g12.4g
B. 24.8g24.8g
C. 0.248g0.248g
D. 49.6g49.6g

Explanation

Solution

The reaction taking place between iodine and nitric acid is not a simple reaction but a redox titration in which the oxidation states of iodine and nitrogen undergo a change as the reaction proceeds. Therefore the n−factorn - factor must be considered during the calculations of mass.

Complete answer:
The reaction given in the equation can be written as follows:
I2+HNO3→HIO3+NO2+H2O{I_2} + HN{O_3} \to HI{O_3} + N{O_2} + {H_2}O
The above reaction is a redox reaction in which iodine is the chemical species getting oxidized and nitrogen is getting reduced. The oxidation and the reduction half reactions can be written as follows:
Oxidation half
I2(0)→2I(+5)+10e−{I_2}(0) \to 2I( + 5) + 10{e^ - }
(The numbers inside round brackets indicate the oxidation states of elements)
Reduction half
HNO3(+5)+e−→NO2(+4)HN{O_3}( + 5) + {e^ - } \to N{O_2}( + 4)
(The oxidation states are associated with the nitrogen atom)
The equivalent mass Meq{M_{eq}} of both the iodine and nitric acid must remain equal during the redox reaction. The formula for calculating equivalent mass is given as follows:
Meq=n−factor×(number of moles){M_{eq}} = n - factor \times ({\text{number of moles)}}
And the number of moles can be calculated using the formula:
number of moles=given massmolar mass{\text{number of moles}} = \dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}
Therefore the number of moles of iodine is:
number of moles of iodine=5g254gmol−1{\text{number of moles of iodine}} = \dfrac{{5g}}{{254gmo{l^{ - 1}}}}
number of moles of nitric acid=x63gmol−1{\text{number of moles of nitric acid}} = \dfrac{x}{{63gmo{l^{ - 1}}}} assuming that the mass of nitric acid is xx
The n−factorn - factor for iodine involved in the oxidation half is 1010.
The n−factorn - factor for reduction involved in the reduction half is 11.
Since the equivalents of iodine and nitric acid are equal, it can be expressed as follows:
Meq(I2)=Meq(HNO3){M_{eq}}({I_2}) = {M_{eq}}(HN{O_3})
10×5g254gmol−1=1×x63gmol−110 \times \dfrac{{5g}}{{254gmo{l^{ - 1}}}} = 1 \times \dfrac{x}{{63gmo{l^{ - 1}}}}
Solving this equation for getting the value of xx :
x=12.4gx = 12.4g

Hence the correct option is (A)

Note:
The n−factorn - factor is the measure of the number of electrons that are involved in a particular reaction per molecule. In an oxidation reaction the n−factorn - factor is the number of electrons lost and in a reduction reaction, it is the number of electrons gained.