Question

$N{a}_{2}C{O}_3+2HCl\to 2NaCl+C{O}_2+H_{2}O$
$106.0$ g of $N{a}_{2}C{O}_3$ reacts with $109.5$ g of $HCl$.

Which of the following is/are correct?

• A. $HCl$ is in excess.
• B. $117.0$ g of $NaCl$ is formed.
• C. The volume of $C{O}_2$ produced at $1$ bar and $273$ K is $22.4$ L.
• D. The volume of $C{O}_2$ produced at $1$ bar and $298$ K is $22.4$ L.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $HCl$ is in excess.
B $117.0$ g of $NaCl$ is formed.
C The volume of $C{O}_2$ produced at $1$ bar and $273$ K is $22.4$ L.
D The volume of $C{O}_2$ produced at $1$ bar and $298$ K is $22.4$ L.

Molecular weight of $N{a}_{2}C{O}_3=106$ g/mol, $HCl=36.5$g/mol and of $NaCl$ $=58.5$ g/mol.

Moles of $N{a}_{2}C{O}_3=\frac{106}{106}=1.0$ mol and moles of $HCl=\frac{109.5}{36.5}=3.0$ mol.

A. Since, for $1$ mol of $N{a}_{2}C{O}_3$, $2$ mol of $HCl$ is required.

So, $HCl$ is in excess $=(3-2)=1.0$ mol

Therefore, $N{a}_{2}C{O}_3$ is the limiting quantity.

B. Weight of $NaCl$ formed $=1\times 2\times 58.5=117.0$ g $NaCl$

C. $1$ mol of $N{a}_{2}C{O}_3=1$ mol of $C{O}_2=22.4L$ at $1$ bar, $273$ K.

D. $1$ mol of $N{a}_{2}C{O}_3=1$ mol of $C{O}_2=22.4$ L at $1$ bar, $298$ K.

Hence, all are correct.