A laboratory assistant prepares a solution of calcium nitrate of volume 250cm³ using Ca(NO3)2.4H2O (s) as the solute: Ca(NO3)2 → Ca2+(aq) + 2NO3-(aq) If the concentration of the nitrate ions in the solution is 0,15 mol·dm-3, calculate the mass of solute that the laboratory assistant used.

The volume of the solution is $250 \text{ cm}^3 = 0.250 \text{ dm}^3$.
Moles of $\text{NO}_3^- = \text{Concentration} \times \text{Volume}$

From the dissociation equation $\text{Ca(NO}_3)_2 \rightarrow \text{Ca}^{2+}(\text{aq}) + 2\text{NO}_3^-(\text{aq})$, we see that 1 mole of $\text{Ca(NO}_3)_2$ produces 2 moles of $\text{NO}_3^-$.
So, moles of $\text{Ca(NO}_3)_2 = \frac{1}{2} \times \text{Moles of NO}_3^-$

Molar mass of $\text{Ca(NO}_3)_2 = 40.08 + 2 \times (14.01 + 3 \times 16.00) = 164.10 \text{ g/mol}$
Molar mass of $4\text{H}_2\text{O} = 4 \times (2 \times 1.01 + 16.00) = 4 \times 18.02 = 72.08 \text{ g/mol}$
Molar mass of $\text{Ca(NO}_3)_2 \cdot 4\text{H}_2\text{O} = 164.10 + 72.08 = 236.18 \text{ g/mol}$

Mass = Moles $\times$ Molar mass

Rounding to three significant figures:

Question 4.2.1:
A limiting reagent is the reactant that is completely consumed in a chemical reaction, thereby determining the maximum amount of product that can be formed.

Question 4.2.2:
To determine the limiting reagent, we compare the mole ratio of reactants used to the stoichiometric mole ratio from the balanced equation.
The balanced equation is: $2\text{C}_7\text{H}_6\text{O}_3 + \text{C}_4\text{H}_6\text{O}_3 \rightarrow 2\text{C}_9\text{H}_8\text{O}_4 + \text{H}_2\text{O}$

Molar mass of salicylic acid ($\text{C}_7\text{H}_6\text{O}_3$): $7(12.01) + 6(1.01) + 3(16.00) = 84.07 + 6.06 + 48.00 = 138.13 \text{ g/mol}$
Molar mass of acetic acid ($\text{C}_4\text{H}_6\text{O}_3$): $4(12.01) + 6(1.01) + 3(16.00) = 48.04 + 6.06 + 48.00 = 102.10 \text{ g/mol}$

Moles of salicylic acid = $\frac{14 \text{ g}}{138.13 \text{ g/mol}} = 0.10135 \text{ mol}$
Moles of acetic acid = $\frac{10 \text{ g}}{102.10 \text{ g/mol}} = 0.09794 \text{ mol}$

From the balanced equation, 2 moles of salicylic acid react with 1 mole of acetic acid.
If all $0.10135 \text{ mol}$ of salicylic acid were to react, it would require:
$\text{Moles of acetic acid needed} = 0.10135 \text{ mol C}_7\text{H}_6\text{O}_3 \times \frac{1 \text{ mol C}_4\text{H}_6\text{O}_3}{2 \text{ mol C}_7\text{H}_6\text{O}_3} = 0.050675 \text{ mol C}_4\text{H}_6\text{O}_3$
Since we have $0.09794 \text{ mol}$ of acetic acid, which is more than the $0.050675 \text{ mol}$ needed, salicylic acid is the limiting reagent.

Alternatively, if all $0.09794 \text{ mol}$ of acetic acid were to react, it would require:
$\text{Moles of salicylic acid needed} = 0.09794 \text{ mol C}_4\text{H}_6\text{O}_3 \times \frac{2 \text{ mol C}_7\text{H}_6\text{O}_3}{1 \text{ mol C}_4\text{H}_6\text{O}_3} = 0.19588 \text{ mol C}_7\text{H}_6\text{O}_3$
Since we only have $0.10135 \text{ mol}$ of salicylic acid, which is less than the $0.19588 \text{ mol}$ needed, salicylic acid is the limiting reagent.

Therefore, the limiting reagent is salicylic acid ($\text{C}_7\text{H}_6\text{O}_3$).

Question 4.2.3:
To calculate the percentage yield, we first need to determine the theoretical yield of aspirin based on the limiting reagent.

From the balanced equation, 2 moles of salicylic acid ($\text{C}_7\text{H}_6\text{O}_3$) produce 2 moles of aspirin ($\text{C}_9\text{H}_8\text{O}_4$).
Since salicylic acid is the limiting reagent (from 4.2.2), we use its moles:
$\text{Moles of aspirin} = 0.10135 \text{ mol C}_7\text{H}_6\text{O}_3 \times \frac{2 \text{ mol C}_9\text{H}_8\text{O}_4}{2 \text{ mol C}_7\text{H}_6\text{O}_3} = 0.10135 \text{ mol}$

Molar mass of $\text{C}_9\text{H}_8\text{O}_4 = 9(12.01) + 8(1.01) + 4(16.00) = 108.09 + 8.08 + 64.00 = 180.17 \text{ g/mol}$

Theoretical yield = Moles $\times$ Molar mass

Actual mass of aspirin obtained = $11.5 \text{ g}$

Rounding to one decimal place: