a) Gọi số mol N₂, O₂ trong 6,72l khí A lần lượt là a, b

=> \(\left\{{}\begin{matrix}28a+32b=8,8\\a+b=\dfrac{6,72}{22,4}=0,3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,1\left(mol\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%V_{N_2}=\dfrac{0,2}{0,3}.100\%=66,67\%\\\%V_{O_2}=\dfrac{0,1}{0,3}.100\%=33,33\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{N_2}=\dfrac{28.0,2}{8,8}.100\%=63,64\%\\\%m_{O_2}=\dfrac{32.0,1}{8,8}.100\%=36,36\%\end{matrix}\right.\)

b) 

\(n_A=0,3\left(mol\right)\)

\(\Rightarrow n_{CO_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{CO_2}=0,3.44=13,2\left(g\right)\)

c) 2,2g A có thể tích là 1,68 lít

=> \(V_{H_2}=1,68\left(l\right)\)

\(a.m_{Mg}=0,1.24=2,4\left(g\right)\\ m_{Ca}=0,2.40=8\left(g\right)\\ b.n_{hh}=\dfrac{2,8}{28}+\dfrac{13,2}{44}=0,4\left(mol\right)\\ \Rightarrow V_{hh}=0,4.22,4=8.96\left(l\right)\)

a) \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

\(n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\)

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(n_{Cu}=\dfrac{19,2}{64}=0,3\left(mol\right)\)

b) 

\(m_{CO_2}=44.0,5=22\left(g\right)\)

\(m_{H_2}=1,5.2=3\left(g\right)\)

\(m_{N_2}=2.28=56\left(g\right)\)

\(m_{CuO}=3.80=240\left(g\right)\)

c) \(n_{Cl_2}=\dfrac{14,2}{71}=0,2\left(mol\right)\)

\(n_{H_2}=\dfrac{4,8}{2}=2,4\left(mol\right)\)

\(n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\)

=> n_hh = 0,2 + 2,4 + 0,1 = 2,7 (mol)

=> V_hh = 2,7.22,4 = 60,48(l)

\(a.n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\\ n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ n_{Cu}=\dfrac{19,2}{64}=0,3\left(mol\right)\)

\(b.m_{CO_2}=0,5.44=22\left(g\right)\\ m_{H_2}=1,5.2=3\left(g\right)\\ m_{N_2}=2.28=56\left(g\right)\\ m_{CuO}=3.80=240\left(g\right)\)

\(c.n_{Cl_2}=\dfrac{14,2}{71}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{4,8}{2}=2,4\left(mol\right)\\ n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\\\Rightarrow n_{hh}=0,2+2,4+0,1=2,7\left(mol\right)\\ \Rightarrow V_{hh}=2,7.22,4=60,48\left(l\right)\)

\(a,n_{Mg}=\dfrac{2,4}{24}=0,1(mol); n_{Zn}=\dfrac{32,5}{65}=0,5(mol)\\ n_{Al}=\dfrac{2,7}{27}=0,1(mol); n_{Cu}=\dfrac{19,2}{64}=0,3(mol)\)

\(b,m_{CO_2}=0,5.44=22(g);m_{H_2}=1,5.2=3(g)\\ m_{N_2}=2.28=56(g);m_{CuO}=3.80=240(g)\)

\(c,n_{hh}=n_{Cl_2}+n_{H_2}+n_{O_2}=\dfrac{14,2}{71}+\dfrac{4,8}{2}+\dfrac{3,2}{32}=0,2+2,4+0,1=2,7(mol)\\ V_{hh}=2,7.22,4=60,48(l)\)

a) V_CO2=0,25.22,4=5,6 (l)

n_N2=56:28=2 mol

V_N2=2.22,4=44,8 (l)

V_hh=44,8+5,6=50,4 (l)

b)n_O3=9,6:48=0,2 mol

V_O3=0,2.22,4=4,48 (l)

  n_C2H2=5,2;26=0,2 mol

V_C2H2=0,2.22,4=4,48 (l)

V_hh=4,48+4,48=8,96 (l)