a, gọi VCH4= x (l), VC2H2=y(l)
CH4 +2 O2 -> CO2 +2 H2O (1)
x............2x........x......................(l)
C2H2 +\(\dfrac{5}{2}\) O2-> 2CO2+H2O (2)
y..............\(\dfrac{5}{2}\)y............2y................(l)
từ (1) và (2) ==> \(\left\{{}\begin{matrix}x+y=28\\2x+\dfrac{5}{2}y=67,2\end{matrix}\right.\)
giải ra x= 5,6 (l) ; y=22,4 (l)
==> % VCH4=\(\dfrac{5,6}{28}.100\%=20\%\)
==> % VC2H2= 100%-20%=80%
b, mCa(OH)2 =\(\dfrac{400.25}{100}\)=100 (g)==> n Ca(OH)2 = 1 (mol)
Vco2= x+ 2y = 5,6 +2.22,4=50,4 (l)==> nCO2 = 2,25 (mol)
xét tỉ lệ \(T=\dfrac{nCO2}{nCa\left(OH\right)2}=\dfrac{2,25}{1}\)=2,25 >2 ==> xảy ra pư
............2CO2 + Ca(OH)2 -> Ca(HCO3)2
trc pư......2,25...........1.....................................(mol)
trg pư........2..............1.........................1............(mol)
sau pư........0,25........0........................1..............(mol)
m dd Ca(HCO3)2 = 2,25.44 + 400=499 (g)
m Ca(HCO3)2= 1.162=162 (G)
==> C% Ca(HCO3)2 =\(\dfrac{162}{499}.100\%\)\(\approx\)32,47%
