Ответ : 1..................................................................
Решение:
Из карбида кальция СаС2.
CaC2+2H2O--->Ca(OH)2+C2H2
3C2H2--[Cакт.]-->С6Н6
C6H6+Cl2--[ FeCl3]-->C6H5Cl+HCl
C6H5Cl+NaOH--->C6H5OH+NaCl
C6H5OH+3Br2--[H2O]-->C6H2Br3OH+3HBr
<span>Cu2O (44,5\%) --> Cu2S(40\%) --> CuFeS2(34,5\%)</span>
M (HCOH)= 300*40%/100% = 120г;
m(CH3OH) = p*V=216*0.79=170.64г;
СН3ОН → НСОН + Н2О;
32г CH3OH ----------30г HCOH
xг CH3OH -------120г HCOH
x=128г
w(CH3OH)=128/170.64*100%=75%