Question

Which of the following is/are correct? The following reaction occurs: $N{a}_{2}C{O}_3+2HCl\to 2NaCl+C{O}_2+H_{2}O$
106.0 g of $N{a}_{2}C{O}_3$ reacts with 109.5 g of $HCl$.

• A. $\text{The HCl is in excess.}$
• B. $\text{117.0 g of NaCl is formed.}$
• C. $\text{The volume of}C{O}_2\text{produced at 1 bar and 273 K is 22.7}L$
  
• D. $\text{The volume of}C{O}_2\text{produced at 1 bar and 273 K is 44.5 L}$

Answer 1

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

The correct options are
A $\text{The HCl is in excess.}$
B $\text{117.0 g of NaCl is formed.}$
C $\text{The volume of}C{O}_2\text{produced at 1 bar and 273 K is 22.7}L$


$\text{MW(molecular weight) of}N{a}_{2}C{O}_3=106,\text{MW of}HCl=36.5\text{MW of}NaCl=58.5$
Moles of $N{a}_{2}C{O}_3=\frac{106}{106}=1.0$ mol

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.
$HCl$ is in excess, $(3-2)=1.0mol$
Therefore, $N{a}_{2}C{O}_3$ is the limiting quantity.

b) Weight of $NaCl$ formed
$2\times 58.5=117g$

c) 1 mol of $N{a}_{2}C{O}_3=1$ mol of $C{O}_2=22.7$ L at 1 bar, 273 K.