Copper (II) Sulphate reacts with Barium Chloride. Given the reaction details, calculate the temperature change of the solution.

Convert volumes from $\text{cm}^3$ to $\text{L}$:
$900 \text{ cm}^3 = 0.900 \text{ L}$
$600 \text{ cm}^3 = 0.600 \text{ L}$

Moles of $\text{CuSO}_4$:

Moles of $\text{BaCl}_2$:

From the balanced equation, $\text{CuSO}_4$ and $\text{BaCl}_2$ react in a $1:1$ molar ratio.
Since $0.600 \text{ mol}$ of $\text{BaCl}_2$ is less than $0.900 \text{ mol}$ of $\text{CuSO}_4$, $\text{BaCl}_2$ is the limiting reactant.
Therefore, $0.600 \text{ mol}$ of reaction occurs.

The enthalpy change for the reaction is $\Delta H = -17.7 \text{ kJ/mol}$.

Convert to Joules:

Total volume of solution:

Mass of solution:

Use the formula $q = mc\Delta T$:

Rounding to three significant figures:

The temperature change is $1.69 \text{ }^\circ\text{C}$.