<span>2Fe + O2=2FeO
</span><span>FeO + H2SO4 = H2O + FeSO4
</span>H2O+H2SO4+2FeSO4=Fe2(SO4)3+2<span>H2O
</span><span>Fe2(SO4)3+6NaOH= 2Fe(OH)3 + 3Na2SO4
</span><span>2Fe(OH)3= Fe2O3 + 3H2O
</span><span>Fe2O3 + 3CО = 2Fe + 3CO2 </span>
CH3-CH-CH2-CH2-CH-CH2-CH2-CH2-CH3
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C2H5 C2H5
Дано
m(ppa CHOH) 75 g
W(CHOH) = 40%
m практ(CHOOH) = 40 g
--------------------------
η-?
m(CHOH)= m(ppa CHOH)*W(CHOH) / 100% = 75*40%/100%= 30 g
M(CHOH) = 30 g/mol
n(CHOH)= m/M= 30 / 30 =1 mol
M(CHOOH) = 46 g/mol
n(CHOOH)= m/M= 40/46 = 0. 9 mol
n(CHOH)> n(CHOOH)
CHOH+Cu(OH)2 -->CHOOH+Cu+H2O
n(CHOH)= n(CHOOH) = 1 mol
m(CHOOH)= n*M= 1*46 = 46 g
η= m(практ)/m(теор) * 100% = 40/46*100% = 90%
ответ 90%