дано
m пр(HCOOH) = 30 g
η(HCOOH) = 90%
--------------------
V(CH4)-?
2CH4+3O2--(kat)-->2HCOOH+2H2O
m теор(HCOOH) = 30 * 100% / 90% = 33.33 g
M(HCOOH) = 46 g/mol
n(HCOOH) = m/M = 33.33 / 46 = 0.72 mol
2n(CH4) = 2n(HCOOH) = 0.72 mol
n(CH4) = 0.72 mol
V(CH4) = Vm * n = 22.4 * 0.72 = 16.128L
ответ 16.128 л
Ω=m(в-ва)/m(р-ра)*100% ⇒ m(в-ва)=8*500/100 =40 г
m(р-ра)=500-200 = 300 г (после выпаривания)
ω=40/300*100%=13,33%