JEE Main 2024 · Chemistry · Mole Concept and Stoichiometry · medium

A 5.00 g mixture of NaHCO3 and Na2CO3 loses 0.93 g on strong heating to constant mass. The volume of 1.00 M HCl needed to react completely (to CO2)…

Single correct option

5.00\ \mathrm{g}

mixture of

\mathrm{NaHCO_3}

and

\mathrm{Na_2CO_3}

loses

0.93\ \mathrm{g}

on strong heating to constant mass. The volume of

1.00\ \mathrm{M}

HCl needed to react completely (to

\mathrm{CO_2}

) with the original mixture is nearly (molar masses:

\mathrm{NaHCO_3}=84

\mathrm{Na_2CO_3}=106

)

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Answer: (A) $76.8\ \mathrm{mL}$

Only bicarbonate decomposes:

2\mathrm{NaHCO_3}\to \mathrm{Na_2CO_3}+\mathrm{CO_2}+\mathrm{H_2O}

, and

168\ \mathrm{g}

of bicarbonate loses

62\ \mathrm{g}

. So

\mathrm{NaHCO_3}=0.93\times168/62=2.52\ \mathrm{g}=0.030

mol. Remaining

\mathrm{Na_2CO_3}=5.00-2.52=2.48\ \mathrm{g}=0.0234

mol. HCl consumed:

1

mol per bicarbonate and

2

mol per carbonate, so

n=0.030+2(0.0234)=0.0768

mol, i.e.

76.8\ \mathrm{mL}

1\ \mathrm{M}

HCl. Using only one HCl per carbonate gives

53.4

mL (trap).

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