Bài 1:
a) \(n_{CO}=x;n_{H_2S}=y\)
\(M_X=32.0,95=30,4\rightarrow\dfrac{28x+34y}{x+y}=30,4\rightarrow2,4x=3,6y\rightarrow\dfrac{x}{y}=\dfrac{3,6}{2,4}=\dfrac{3}{2}\)
%VCO=\(\dfrac{3}{3+2}.100=60\%\)
%\(V_{H_2S}=\dfrac{2}{3+2}.100=40\%\)
b) \(V_{CO}=12,32.\dfrac{3}{5}=7,392l\)
\(V_{H_2S}=12,32.\dfrac{2}{5}=4,928l\)
-Gọi \(V_{CO_2}=x\left(lít\right)\)
\(\overline{M}=\dfrac{44x+28.7,392+34.4,928}{x+12,32}=9.2=18\)
Giải ra x\(\approx5,9\left(lít\right)\)
Bài 2:\(n_{O_2}=x\left(mol\right)\); \(n_{O_3}=y\left(mol\right)\)
\(\overline{M}=\dfrac{32x+48y}{x+y}=18.2=36\rightarrow4x=12y\rightarrow x=3y\)
Ngoài ra 32x+48y=30
Giải 2 phương trình trên ta được: x=\(\dfrac{5}{8}\) và y=\(\dfrac{5}{24}\)
\(m_{O_2}=32x=32.\dfrac{5}{8}=20\left(gam\right)\)
\(m_{O_3}=30-20=10\left(gam\right)\)