[dorask]

In a heat isolated system, in the temperature of 20C, which contains solution of 50gr with 4% (mass) NaOH and a solution of 50gr with 1.825% HCl. After the reaction the temperature reached 23.4C. Afterwards a 70gr solution, in 20C, of 3.5% H2SO4 was added to the main isolated solution.
1.What's the final temperature of the solution? (specific heat capacity of the solution – Cp=4.19J/g*Celzius) (Q=C*m*delta-t, delta-H=-Q/n) – answer:24C
2. calculate the mass of the remaining dry solid after evaporation of water from the solution. Answer: 3.55gr

In question 1 can you please tell me how could it be that there a specific heat capcity of a solution. Please show steps and what formulas you used to solve it.
In question two my answer was 3.55gr of Na2SO4 and 0.025gr of NaCl. Where were I wrong?

[Borek]

Please read forum rules. You have to show your attempts at solving the question to receive help. This is a forum policy.

All you need is a stoichiometry of reactions and two equations that you already listed.

In question 1 can you please tell me how could it be that there a specific heat capcity of a solution.

Specific heat of solution what? That there exists a specific heat capacity of a solution?

In question two my answer was 3.55gr of Na2SO4 and 0.025gr of NaCl. Where were I wrong?

Everywhere. Both numbers are wrong. There is no 3.55 g of Na₂SO₄ (less than that), there is no 0.025 g of NaOH (although 0.025 is a number that is related to the result).

[dorask]

look in question number 1 i just couldn't realize how to solve it. this material is quite new to me and i don't know from where to begin. I didn't realize how could it be that a heat capacity of a solution exists because the solution has water in it, NaOh, HCl and the n you add it H2SO4 so how could it stay the same heat capacity???

it question number 2:
the first reaction - NaOH + HCl --> NaCl  + H2O
from the question there are 2gr of NaOH and 0.05 moles of it. HCl is 0.9125gr which is 0.025 moles. so there is 0.025 NaCl left as much as NaOh.
the second : 2NaOH + H2SO4 --> Na2SO4 + 2H2O
there are 0.025 moles left of NaOH and also 2.45gr of H2SO4 so the fully react and there is 0.025 moles of Na2So4 ,Mw=142 -> m = 3.55gr. no can you please explain me where i was wrong i can't see it.. (and also could you explain the first que.)

thank you very much i really appriciate it

[Borek]

I didn't realize how could it be that a heat capacity of a solution exists

Everything has a heat capacity. You heat it, you check by how much the temperature increased, you calculate the ratio - that's the heat capacity.

because the solution has water in it, NaOh, HCl and the n you add it H2SO4 so how could it stay the same heat capacity???

It should change and not be identical to water, but usually we assume for a diluted solutions it is not much different.

the first reaction - NaOH + HCl --> NaCl  + H2O
from the question there are 2gr of NaOH and 0.05 moles of it. HCl is 0.9125gr which is 0.025 moles. so there is 0.025 NaCl left as much as NaOh.

0.025 of WHAT (name the unit) of NaCl?

the second : 2NaOH + H2SO4 --> Na2SO4 + 2H2O
there are 0.025 moles left of NaOH and also 2.45gr of H2SO4 so the fully react and there is 0.025 moles of Na2So4 ,Mw=142 -> m = 3.55gr

This is a limiting reagent question.

First question - use information given to calculate enthalpy of neutralization (H⁺+OH^- H₂O). How many moles of H⁺ were neutralized?