Рассчитаем избыток
mMg(NO3)2=250·0.1=25g
nMg(NO3)2=25/148=0.17mol
mKOH=150·0.1=15g
nKOH=15/56=0.27mol - izbytok
25g xg
Mg(NO3)2+2KOH=Mg(OH)2↓+2KNO3
148g 58g
x=25·58/148=9.8g
M(CxHy) = DH2*MH2 = 56*2 =112 г/моль
n(C) = n(CO2) = m/M = 35,2/44 = 0,8 моль
n(H) = 2n(H2O) = m/M *2 = 16,2/18*2 = 1,8 моль
m(C) = A*n = 0,8*12 = 9,6 г
m(H) = 1,8 г
11,4-9,6-1,8 = 0
n(C):n(H) = 0,8:1,8 = 1:2
CH2
M(CH2) = 14 г/моль
112/14 = 8
C8H16
<span>S + Fe = FeS
</span>FeS + H2O = FeO + H2S
<span>2 H2S + 3 O2 = 2 H2O + 2 SO<span>2
</span></span><span>2 SO2 + O2 = 2 SO<span>3
</span></span><span>SO3 + H2O = H2SO<span>4
</span></span><span>2 H2SO4 = 2 H2O + O2 + 2 SO<span>2</span></span>