Solving Acid-Base Reactions: NaOH + HCl → NaCl + H2O

[Glen Maverick]

Homework Statement

NaOH(aq) + HCl(aq) ------> NaCl(aq) + H2O(l)

50mL of 1.0M HCl and 50ml 1.0M NaOH was mixed. Total heat is 2826J. If the density of 0.5M NaCl produced in this reaction is 1.02g/mL:
a) What is the mass of the solution?
b) Moles of water formed
c) Moles of NaCl formed
d) Molar heat of reaction

Homework Equations

The Attempt at a Solution

a) 1.02g/ml x 100ml = 102 g
b) 50ml x 0.001 x 1M =0.05 moles of Water
d) 2826/0.05 = 56.52kJ

Iam very confused about c). Why is there "0.5M"in the question? Do I have to use that molarity during calculation?

 

[Borek]

I am not sure what is so confusing? If you start with amounts of reagents given, you calculate number of moles of NaCl produced, and you check what is the final volume, you will find 0.5 M is just the final concentration of NaCl. I suppose teacher will prefer you to calculate amount of product from the stoichiometry, not from the final concentration, but both ways give correct result.

 

[Glen Maverick]

I get it. :) What I am also confused about is whether the mass of the solution os 102g or 100g. TA said we ar multiplying the dnesity of NaCl to 100ml(50ml+50ml) for finding heat gained by solution. according to the lab manual, the calculation is saying 102 g. However, I think 100g is the mass of the solution.

 

[Borek]

What volume? What density? What mass then?

 

[DeeNos]

In order to solve this problem, it is important to first understand the chemical reaction that is occurring. In this case, we have a neutralization reaction between a strong acid (HCl) and a strong base (NaOH). This means that the products of the reaction will be a salt (NaCl) and water (H2O).

Now, let's address the questions one by one:

a) The mass of the solution can be calculated by first finding the total volume of the solution. Since we are mixing equal volumes of 50mL of 1.0M HCl and 50mL of 1.0M NaOH, the total volume will be 100mL. Using the density given, we can calculate the mass of the solution to be 102g.

b) The moles of water formed can be calculated using the balanced chemical equation. For every 1 mole of HCl, 1 mole of water is formed. We have 0.05 moles of HCl in the solution, so we will also have 0.05 moles of water formed.

c) The moles of NaCl formed can be calculated using the same method as in part b. For every 1 mole of HCl, 1 mole of NaCl is formed. Therefore, we will have 0.05 moles of NaCl formed.

d) The molar heat of the reaction can be calculated by dividing the total heat (2826J) by the moles of water formed (0.05 moles). This gives us a molar heat of reaction of 56.52 kJ/mol.

As for the confusion about the 0.5M NaCl, it is not directly used in the calculations. It is simply given to us as the molarity of the NaCl produced in the reaction. We do not need it to solve the problem, but it is useful information to know. I hope this helps to clarify any confusion.