Question

# The volume of $CO_2$ (in ${ cm }^{ 3 }$) liberated at S.T.P. when 1.06g of anhydrous sodium carbonate is treated with an excess of dilute HCl is:A- 112B- 224C- 56D- 2240

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Hint: Try to figure out the reaction between sodium carbonate and dilute HCl. Think about whether HCl is stronger acid or carbonic acid is stronger. Think about the volume of 1mole of the gas at S.T.P. By balancing reaction and then calculate the number of moles of $CO_2$. Find the volume of gas liberated.

We know that anhydrous sodium carbonate reacts with dilute HCl. This reaction is because Hydrochloric acid is a stronger acid than carbonic acid. Whenever there is a reaction between salt and acid which is a stronger acid than the conjugate acid of the anion of salt then anion of salt is replaced by a conjugate base of strong acid and conjugate acid of the anion of salt is formed. Let HA, HB be acids where HB is a stronger acid than HA. Let COH be a base. CA be salt of HA and COH.

$CA\quad +\quad HB\quad \rightarrow \quad CB\quad +\quad HA$

A similar reaction happens between HCl and sodium carbonate. But some energy will be released when NaCl is formed this energy is released as heat which will be utilized by carbonic acid and it dissociates into carbon dioxide. That’s why $CO_2$ is liberated.

$Na_{ 2 }CO_{ 3 }\quad +\quad 2HCl\quad →\quad 2NaCl\quad +\quad H_{ 2 }CO_{ 3 }$
$H_{ 2 }CO_{ 3 }\quad +\quad heat\quad →\quad CO_{ 2 }\quad +\quad H_{ 2 }O$

We know that 1 mole of any gas has a volume of 22.4L at S.T.P conditions.
Moles of $Na_{ 2 }CO_{ 3 }$ = moles of $H_{ 2 }CO_{ 3 }$ = moles of $CO_2$
Moles of $Na_{ 2 }CO_{ 3 }$ = $\frac { weight\quad of\quad Na_{ 2 }CO_{ 3 } }{ molecular\quad weight\quad of\quad Na_{ 2 }CO_{ 3 } }$

$Molecular\quad weight\quad of\quad Na_{ 2 }CO_{ 3 }$ = 106 $g\quad mol^{ -1 }$
$weight\quad of\quad Na_{ 2 }CO_{ 3 }\quad =\quad 1.06\quad g$
Moles of $Na_{ 2 }CO_{ 3 }$ = $\frac { 1.06\quad g }{ 106\quad g\quad mol^{ -1 } } \quad =\quad 0.01\quad mol$
Moles of $CO_2$= 0.01
1mol of any gas has 22.4 L.
Volume of $CO_{ 2 }\quad =\quad 0.01×22.4L\quad =\quad 0.224L\quad =\quad 224mL\quad =\quad 224cm^{ 3 }.$
Hence, the correct answer is option B.

Note:
Heat will be released when sodium chloride is formed from sodium carbonate because sodium chloride has more lattice energy per equivalent compared to that of sodium carbonate and the difference is released as heat energy.