Here’s <em>one of many</em> possibilities.
We <em>MUST</em> know the heat of reaction, Δ<em>H</em>, <em>before we start
</em>.
Let’s <em>assume</em> that the reaction is
A + B → Products; Δ<em>H</em> = -80 kJ·mol⁻¹
Let’s <em>assume</em> that you want to get <em>100 mL</em> of solution that warms from <em>25 °C to 50 °C</em>.
<em>For the solution
</em>
<em>q = mc</em>Δ<em>T</em> = 100 g × 4.184 J·K⁻¹mol⁻¹ × 25 K = 10 460 J = 10.46 kJ
The reaction must supply 10.46 kJ.
<em>For the reaction</em>
<em>n</em>Δ<em>H</em> = 10.46 kJ
<em>n</em> = 10.46 kJ/80 kJ = 0.131 mol
So, you need <em>0.131 mol A</em> and 0.131 mol B.
<em>Assume</em> you are using the <em>3 mol·L⁻¹ </em>solutions.
Then
<em>V</em> = 0.131 mol × (1 L/3 mol) = 0.0436 L = 46.6 mL
You need 46.6 mL of 3 mol·L⁻¹ A + 46.6 mL of 3 mol·L⁻¹ B.
Add 3.4 mL distilled water to 46.6 mL of 3 mol·L⁻¹ A to make 50 mL of A.
Add 3.4 mL distilled water to<em> </em>46.6 mL of 3 mol·L⁻¹ B to make 50 mL of B.
Mix the two solutions, and you will have 100 mL of a solution at 50 °C
.
