Question

In the Solvay process for producing sodium carbonate $\left( Na _{2} CO _{3}\right),$ the following reactions occur in sequence
$NH _{3}+ CO _{2}+ H _{2} O \longrightarrow NH _{4} HCO _{3}$
$NH _{4} HCO _{3}+ NaCl \longrightarrow NH _{4} Cl + NaHCO _{3}$
$2 NaHCO _{3} \stackrel{\Delta}{\longrightarrow } Na _{2} CO _{3}+ H _{2} O + CO _{2}$
How much of $Na _{2} CO _{3}$ would be produced per $kg$ of $NH _{3}$ used if the process were $100 \%$ efficient?

• A. 3.12 kg
• B. 1.00 kg
• C. 2.00 kg
• D. 0.56 kg

Answer 1

Answer: (A)

Given reaction is not balanced

Based on stoichiometry

$1\,mol \,Na _{2} CO _{3}$ is from $2 \,mol \,NaHCO _{3}$

$2 \,mol\, NaHCO _{3}$ is from $2\, mol \,NH _{4} HCO _{3}$

$2\, mol \,NH _{4} HCO _{3}$ is from $2\, mol\, NH _{3}$

Thus, $1 \,mol\, Na _{2} CO _{3}$ is from $=2 \,mol \,NH _{3}$

$106 \,g\, Na _{2} CO _{3}$ is from $=2 \times 17\, g\, NH _{3}$

$34\, g\, NH _{3}$ gives $=106 \,g \,Na _{2} CO _{3}$

Thus, $1000 \,g \,NH _{3}$ gives $=\frac{106 \times 1000}{34} \,g\, Na _{2} CO _{3}$

$=3118 \,g\, Na _{2} CO _{3}$

$=3.118 \,kg\, Na _{2} CO _{3}$