Question

physical science/control test 5 nw/september 2025
question 2
in an experiment, shipilo, a grade 11 learner added 1.5 g of sodium carbonate (na₂co₃) to hydrochloric acid (hcl). a volume of 306 cm³ of carbon dioxide gas was formed and collected under standard pressure at room temperature. take the molar gas volume at room temperature (vm) as 24.43 dm³.
the unbalanced equation for the reaction is:
na₂co₃(s) + hcl(aq) → nacl(aq) + h₂o(l) + co₂(g)
2.1 define the term one mole of a substance. (2)
2.2 balance the equation for the reaction. (2)
2.3 calculate the mass of sodium carbonate that reacted. (7)
2.4 calculate the percentage of sodium carbonate in excess. (2)
13

Explanation:

2.1

One mole of a substance is the amount of that substance that contains as many elementary entities (such as atoms, molecules, or ions) as there are atoms in exactly 12 g of carbon - 12.

2.2

The balanced equation is $Na_2CO_3(s)+2HCl(aq)
ightarrow 2NaCl(aq)+H_2O(l)+CO_2(g)$

2.3

Step1: Calculate the number of moles of $CO_2$

The volume of $CO_2$ is $V = 306\ cm^3=0.306\ dm^3$. Given the molar gas volume $V_m = 24.43\ dm^3/mol$. The number of moles of $CO_2$, $n(CO_2)=\frac{V}{V_m}=\frac{0.306}{24.43}\ mol$

Step2: Determine the mole - ratio between $Na_2CO_3$ and $CO_2$

From the balanced equation, the mole - ratio $n(Na_2CO_3):n(CO_2)=1:1$. So $n(Na_2CO_3)=n(CO_2)=\frac{0.306}{24.43}\ mol$

Step3: Calculate the molar mass of $Na_2CO_3$

The molar mass of $Na_2CO_3$, $M(Na_2CO_3)=(2\times23 + 12+3\times16)\ g/mol=106\ g/mol$

Step4: Calculate the mass of $Na_2CO_3$ that reacted

$m(Na_2CO_3)=n(Na_2CO_3)\times M(Na_2CO_3)=\frac{0.306}{24.43}\times106\ g\approx1.32\ g$

2.4

Step1: Calculate the mass of excess $Na_2CO_3$

The initial mass of $Na_2CO_3$ is $m_{initial}=1.5\ g$, and the mass that reacted is $m_{reacted}\approx1.32\ g$. So the mass of excess $Na_2CO_3$, $m_{excess}=1.5 - 1.32 = 0.18\ g$

Step2: Calculate the percentage of excess $Na_2CO_3$

The percentage of excess $Na_2CO_3=\frac{m_{excess}}{m_{initial}}\times100=\frac{0.18}{1.5}\times100 = 12\%$

Answer:

2.1: One mole of a substance is the amount of that substance containing as many elementary entities as there are atoms in 12 g of carbon - 12.
2.2: $Na_2CO_3(s)+2HCl(aq)
ightarrow 2NaCl(aq)+H_2O(l)+CO_2(g)$
2.3: $1.32\ g$
2.4: $12\%$