Help me solve this calorimetry problem please!

Question

One beaker contains 29 mL of 1.60 M HCl, and a second beaker contains 50 mL of 0.40 M NaOH. Both solutions were initially at 20 °C. We pour both beakers into a large insulated container which contains 6 g of ice, initially at -60 °C. Calculate the final temperature of the resulting solution. Note that the heat of neutralization is -57.3 kJ/mol.

Answer 1

This is quite an involved question. Let's break it down step by step. First, we'll start with our balanced equation

NaOH + HCl ----> NaCl + H₂O

The ∆H for this reaction is -57.3 kJ/mol meaning exothermic or heat is released. Once we have that we have to calculate the mols of reactant and our limiting reagent. Remember M=molarity and the formula for Molarity is mols/Liters.

We would set up for HCl--> 1.6M = x mols/0.029 L and find our x to be 0.0464 mols HCl reacted

We would set up for NaOH--> 0.4M = x mols/0.050 L and find our x to be 0.02 mols NaOH reacted

Because this is a 1 to 1 ratio (HCl-NaOH) our limiting reagent is NaOH so all of our NaOH will be reacted.

Additionally for every 1 mol of NaOH reacted we have 57.3 kJ of heat being released (heating up the water)

Therefore if we only have 0.02 mol NaOH then we'll have 1.146 kJ or 1146 Joules of heat released. This is our q.

We'll assume that we have an ideal calorimeter and all the heat released gets directly transferred to the water so q (released by the reaction) = q(gained by the water in the calorimeter).

remember q=mc∆t

It was pointed out that if we may have enough energy for the ice to change to water. But before we can do that we have to first bring the Ice up to 0 degrees C from -60. that q=mc∆t we solve for q using

m (6 g), c (for ice 2.03 J/(g*C), and 60 for ∆t.

We find our q to bring the ice up to 0C is 730.8 Joules. which means we only have 415.2 Joules of energy left. The heat of fusion for water is 334 J/g. Heat of fusion is the energy required to melt all the ice to liquid water

We don't have enough to heat all 6 grams of ice to water then the solution would remain at 0C

I hope this helps