Question

Complete the following equations:

$2{\mathrm{NaHCO}}_3+{\mathrm{H}}_2{\mathrm{SO}}_4\to \mathrm{\_\_\_\_\_\_\_\_}+2{\mathrm{H}}_2\mathrm{O}+\mathrm{\_\_\_\_\_\_}$

• A. ${Na}_2{SO}_4$
• B. $2{CO}_2$

Answer 1

Reaction summary - Reaction of sodium bicarbonate $\left({\mathrm{NaHCO}}_3\right)$ and sulfuric acid $\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$

When Sodium bicarbonate reacts with Sulfuric acid, Sodium sulfate $\left({\mathrm{Na}}_2{\mathrm{SO}}_4\right)$ is formed along with water $\left({\mathrm{H}}_2\mathrm{O}\right)$ and Carbon dioxide $\left({\mathrm{CO}}_2\right)$.

The skeletal reaction is as follows:

${\mathrm{NaHCO}}_3\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to {\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)$

• There are $2$ sodium $\left(\mathrm{Na}\right)$ atoms on the product side, while on the reactant side, there is $1$only. To balance sodium atoms, a coefficient of $2$ must be added in front of ${\mathrm{NaHCO}}_3$.
• The number of hydrogen $\left(\mathrm{H}\right)$ on both sides of the reactions is unbalanced. To balance the number of hydrogen, a coefficient of $2$ must be added in front of ${\mathrm{H}}_2\mathrm{O}$.
• However, by this carbon$\left(\text{C}\right)$ and oxygen $\left(\mathrm{O}\right)$ count on both sides of the reaction become unbalanced. To balance the number of carbon and oxygen, a coefficient of $2$ must be added in front of ${\mathrm{CO}}_2$.

The balanced chemical reaction is as follows:

$2{\mathrm{NaHCO}}_3\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to {\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+2{\mathrm{CO}}_2\left(\mathrm{g}\right)$

Therefore, the complete equation is $2{\mathrm{NaHCO}}_3+{\mathrm{H}}_2{\mathrm{SO}}_4\to \underline{{\mathrm{Na}}_2{\mathrm{SO}}_4}+2{\mathrm{H}}_2\mathrm{O}+\underline{2{\mathrm{CO}}_2}$