Question

The equation given below related to the manufacture of sodium carbonate [molecular weight of $N{{a}_{2}}C{{O}_{3}}$ = 106].
a. $NaCl+N{{H}_{3}}+C{{O}_{2}}+{{H}_{2}}O\to NaHC{{O}_{3}}+N{{H}_{4}}Cl$
b. $2NaHC{{O}_{3}}\to N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O+C{{O}_{2}}$
Questions (i) and (ii) are based on the production of 21.2 g of sodium carbonate.
(i) What mass of sodium hydrogen carbonate must be heated to give 21.2 g of sodium carbonate. [molecular weight of $NaHC{{O}_{3}}$ = 84]?
[Type only the numerical value of the mass, assuming that its units is grams]
(ii) To produce the mass of sodium hydrogen carbonate calculated in (i), what volume of carbon dioxide, measured at STP, would be required?
[Type in only the numerical value of the volume, assuming that its unit in liters]

Answer 1

Sodium carbonate is going to prepare from two moles of sodium hydrogen carbonate or sodium bicarbonate and sodium bicarbonate is going to prepare from the reaction of sodium chloride with ammonia, carbon dioxide and water.

Complete step by step answer:
- The given chemical reactions in the question to prepare sodium hydrogen carbonate and sodium bicarbonate are as follows.
a. $NaCl+N{{H}_{3}}+C{{O}_{2}}+{{H}_{2}}O\to NaHC{{O}_{3}}+N{{H}_{4}}Cl$
b. $2NaHC{{O}_{3}}\to N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O+C{{O}_{2}}$
(i)
- From the above chemical equation (b) we can say easily that 2 moles of sodium hydrogen carbonate gives one moles of sodium carbonate.
$\underset{2\times 84}{\mathop{2NaHC{{O}_{3}}}}\,\equiv \underset{\text{106}}{\mathop{N{{a}_{2}}C{{O}_{3}}}}\,$
- Means to get 106 g of sodium carbonate we have to heat 168 g of sodium hydrogen carbonate.
- Therefore to obtain 21.2 g of sodium carbonate can be obtained by heating how much amount of sodium bicarbonate.
- To obtain 21.2 g of sodium carbonate can be obtained by heating sodium bicarbonate

& =\dfrac{168}{106}\times 21.2 \\\ & =33.6g \\\ \end{aligned}$$ **\- The mass of sodium hydrogen carbonate heated to obtain 21.2 g sodium carbonate is 33.6 g.** (ii) - From the above chemical reaction (a) we can say that 1 mole of sodium hydrogen carbonate is obtained as reaction 1 mole of sodium chloride with one mole of carbon dioxide. \- So, $$\underset{84g\text{ }}{\mathop{NaHC{{O}_{3}}}}\,\equiv \underset{\text{22}\text{.4 litre}}{\mathop{C{{O}_{2}}}}\,$$ \- 84 g of sodium hydrogen carbonate at STP is equivalent to 22.4 liter of carbon dioxide. \- Therefore 33.6 g of sodium hydrogen carbonate at STP is equal to $$\begin{aligned} & =\dfrac{22.4}{84}\times 33.6 \\\ & =8.96liter \\\ \end{aligned}$$ **\- The produce 33.6 g of sodium hydrogen carbonate the volume of carbon dioxide required is 8.96 liter.** **Note:** Sodium bicarbonate is also called baking soda. Because whenever we are going to heat the sodium bicarbonate it releases carbon dioxide as shown in the equation below. $2NaHC{{O}_{3}}\to N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O+C{{O}_{2}}$ The formed carbon dioxide causes lightens in weight and softens the material which is under the baking process.