[Rutherford]

In a reactor CaO and C are added in stoichiometric amounts. A sample was taken from the reactor and after careful dissolution in HCl (a gas was intensively made, c(HCl)=0.4M, V=750ml)), the solution was filtrated (mass of the non-dissolved residue was 0.879g), and water was added until the volume became 1dm³. For the titration of 20cm³ of this solution, 9.86ml of NaOH (c=0.1M) were spent. What was the yield (in percents)of the reaction that took place in the reactor?

CaO+C Ca+CO
Then:
CaO+2HCl CaCl₂+H₂O
Ca+2HCl CaCl₂+H₂
The non-dissolved residue should be C (n=0.879/12=0.07325mol, meaning that there was 0.07325mol of CaO). The remaining HCl was titrated. n of HCl titrated is 0.00986*0.1, in the 1dm³ there was 50 times more HCl, so the amount of HCl that reacted with Ca and CaO is: n=0.75*0.4-0.00986*0.1*50=0.2507mol of HCl.
What to do now? The HCl reacted with (0.07325-x)mol of CaO, and with xmol of Ca, meaning that HCl reacted with 0.07325mol of Ca²⁺, but that doesn't help me.

[DrCMS]

Yes the HCl reacts with both Ca and CaO but only one of those reactions produces a gas.

[Rutherford]

If it says that a gas is made, it doesn't mean that CaO is not reacting with water. That's is the problem.

[Borek]

If CaO and C are in stoichiometric amounts, solid residue can't be C.

I am not convinced you can reduce CaO with C to metallic Ca, but I can't check it right now in my usual sources.

[AWK]

CaO+C  Ca+CO

This is not a method of production Ca. CaC₂ is obtained in this method.
Ca is obtained from CaO and Al. Vapours of Ca are condensed and Ca are practically pure.
CaC2 reacts with water or acid and forms gas (acetylene).

[Rutherford]

Okay, I did it this way:
CaO+2C CaC₂+1/2O₂
CaC₂+2H₂O C₂H₂+Ca(OH)₂
The CaO that left from the first reaction reacts with water, too:
CaO+H₂O Ca(OH)₂
Both Ca(OH)₂ react with HCl (whose n is the same I got previously) n=0.2507mol
I can't think of something else to be the residue other that non-reacted C, so n(C) that didn't react was n=0.879/12=0.07325mol, the amount of non-reacted CaO was two times smaller n=0.036625mol.
Then I get the final equation:
0.2507(amount of HCl that reacted with Ca(OH)₂)=2*0.036625(amount of Ca(OH)₂ that was made from the non-reacted CaO)+2x(amount of Ca(OH)₂ made from CaC₂)
x=0.033725mol, this is also the number of moles of CaC₂, the initial number of moles of CaO was 0.033725(moles that reacted)+0.036625(moles that didn't react)=0.07035mol
yield=0.033725/0.07035*100=48% much smaller that the right answer  .

[AWK]

CaO+3C = CaC₂+CO
This is an equilibrium reaction, hence both CaO and C was left, but after addition of  diluted hydrochloric acid only C is a residue. Since stoichiometric amount of reagents were used  - from unreacted C you can calculate moles (and mass) of unreacted CaO, and  from back titration total CaO used. Finally you can obtain yield of this reaction.

[Rutherford]

I did everything you said now in my previous post, I only missed the reaction  . Anyway, thanks for your help, without it I couldn't think of the right reaction. I got the right result now, which is 80.5%.