Question

Question: 2.10 g mixture of \[{\text{NaHC}}{{\text{O}}_{\text{3}}}\] and ${\text{KCl}}{{\text{O}}_{\text{3}}}$ requires 100 mL of 0.1N HCl for complete reaction. If x grams amount of residue that would be obtained on strongly heating 2.20 g of the same mixture then, determine the value of 1000x.
A. 1358
B. 1300
C. 1380
D. None of the above

Answer 1

Hint: Assume one of the reactants weigh ‘a’ gram from the total of 2.10 g, and recall the concepts of stoichiometry and stoichiometric calculations, and to get these values write the balanced chemical reaction of heating of \[{\text{NaHC}}{{\text{O}}_{\text{3}}}\] and ${\text{KCl}}{{\text{O}}_{\text{3}}}$

Complete Step by step answer: As mentioned in hint let’s assume the weight of \[{\text{NaHC}}{{\text{O}}_{\text{3}}}\] as ‘a’,
$ \Rightarrow $ weight of ${\text{KCl}}{{\text{O}}_{\text{3}}}$ = 2.10 $ - $ a (since the total weight of the mixture is 2.10 g)
Since only \[{\text{NaHC}}{{\text{O}}_{\text{3}}}\] will react with the 0.1N HCl solution, we can calculate its mass using the formula of normality,
$ \Rightarrow $ ${\text{Normality(N) = }}\dfrac{{{\text{No}}{\text{. of gram equivalent of solute }}}}{{{\text{Volume of solution(L)}}}}$
$ \Rightarrow $${\text{No}}{\text{. gram equivalent of NaHC}}{{\text{O}}_3}{\text{ = N }} \times {\text{ Volume of solution(L)}}$
$ \Rightarrow $${\text{No}}{\text{. gram equivalent of NaHC}}{{\text{O}}_3}$ = 0.1 $ \times {\text{ }}100 \times {10^{ - 3}}$ (we have changed the unit ml to L by multiplying 1000 with it. Since 1L = 1000ml)
$ \Rightarrow $${\text{No}}{\text{. gram equivalent of NaHC}}{{\text{O}}_3}$ = ${10^{ - 2}}$
We can use the gram equivalent weight of \[{\text{NaHC}}{{\text{O}}_{\text{3}}}\] to find out its mass, using the formula,
$ \Rightarrow $${\text{No}}{\text{. gram equivalent of NaHC}}{{\text{O}}_3} = {\text{ }}\dfrac{{{\text{weight of NaHC}}{{\text{O}}_3}}}{{{\text{Equivalent weight NaHC}}{{\text{O}}_3}}}$
$ \Rightarrow $${\text{weight of NaHC}}{{\text{O}}_3}{\text{ = No}}{\text{. gram equivalent of NaHC}}{{\text{O}}_3}{\text{ }} \times {\text{ Equivalent weight NaHC}}{{\text{O}}_3}$
$ \Rightarrow $${\text{weight of NaHC}}{{\text{O}}_3}$ = ${10^{ - 2}}$$ \times $ 84 (equivalent weight of \[{\text{NaHC}}{{\text{O}}_{\text{3}}}\] is 84g)
$ \Rightarrow $${\text{weight of NaHC}}{{\text{O}}_3}$ = 0.84 g
$ \Rightarrow $ a = 0.84g
Now we can calculate the weight of ${\text{KCl}}{{\text{O}}_{\text{3}}}$,
$ \Rightarrow $ weight of ${\text{KCl}}{{\text{O}}_{\text{3}}}$ = 2.10 $ - $ a
$ \Rightarrow $ weight of ${\text{KCl}}{{\text{O}}_{\text{3}}}$ = 2.10 $ - $ 0.84
$ \Rightarrow $ weight of ${\text{KCl}}{{\text{O}}_{\text{3}}}$ = 1.26g
It is time to write the balanced chemical equation for the heating of ${\text{KCl}}{{\text{O}}_{\text{3}}}$ and \[{\text{NaHC}}{{\text{O}}_{\text{3}}}\]
iii. ${\text{2NaHC}}{{\text{O}}_{\text{3}}}\xrightarrow{\Delta }{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O }} \uparrow {\text{ + C}}{{\text{O}}_{\text{2}}} \uparrow $
iv. ${\text{2KCl}}{{\text{O}}_{\text{3}}}\xrightarrow{\Delta }{\text{2KCl + 3}}{{\text{O}}_{\text{2}}} \uparrow $
From the above chemical reaction, we can conclude that
2 moles of \[{\text{NaHC}}{{\text{O}}_{\text{3}}}\] and 2 moles of ${\text{KCl}}{{\text{O}}_{\text{3}}}$ are being consumed, also the residue are ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ and KCl in the reaction since we are heating, the water molecules will get evaporated and carbon dioxide and oxygen gas will be liberated.
Now, the weight of 2 moles of \[{\text{NaHC}}{{\text{O}}_{\text{3}}}\], 2 mole of ${\text{KCl}}{{\text{O}}_{\text{3}}}$, 2 moles of KCl and and 1 mole of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ are;
$ \Rightarrow $ weight of 2 moles of \[{\text{NaHC}}{{\text{O}}_{\text{3}}}\] = 2 $ \times $ 84.007 = 168.01g (molecular mass of \[{\text{NaHC}}{{\text{O}}_{\text{3}}}\] = 84)
$ \Rightarrow $ weight of 2 moles of ${\text{KCl}}{{\text{O}}_{\text{3}}}$ = 122.5 $ \times $ 2 = 245g
$ \Rightarrow $ weight of 2 moles of KCl = 74.5$ \times $2 = 149g
$ \Rightarrow $ weight of 1 mole of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ = 106
$\therefore $The weight of residue from 2.1 g of the mixture = $\dfrac{{{\text{weight of 1 mole of N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}}}{{{\text{weight of 2 moles of NaHC}}{{\text{O}}_3}}} \times 0.84{\text{ + }}\dfrac{{{\text{weight of 2 moles of KCl}}}}{{{\text{weight of 2 moles of KCl}}{{\text{O}}_3}}} \times 1.26$
$ \Rightarrow $ $\dfrac{{106}}{{168.01}} \times 0.84{\text{ }} + {\text{ }}\dfrac{{149}}{{245}} \times 1.26$ = 0.53 + 0.766 = 1.296
$\therefore $Weight of residue from 2.2g of the mixture = $\dfrac{{1.296}}{{2.1}} \times 2.2$ = 1.358g
$\therefore $ the value of x = 1.358
$ \Rightarrow $ 1000x = 1358

Hence, the correct answer is option is (A) i.e., 1358.

Note: One may end their solution after finding the weight of residue in 2.10 g of the solution mixture assuming it to the final answer, however, the question asks for the weight of residue in 2.2 g solution of the same mixture. Hence, carefully understanding the question is necessary.