a)
Gọi n O2= a(mol) ; n N2 = b(mol)
Coi n X = 1(mol) ; M X = 14,8.2 = 29,6 (g/mol)
Ta có :
n X = a + b = 1(mol)
m X = 32a + 28b = 29,6.1 = 29,6(gam)
Suy ra : a =0,4 ; b = 0,6
Ta có :
V O2 : V N2 = 0,4 : 0,6 = 2 : 3
b)
%m O2 = 0,4.32/29,6 .100% = 43,24%
%m N2 = 100%- 43,24% = 56,76%
\(GS:n_X=1\left(mol\right)\)
\(n_{O_2}=a\left(mol\right),n_{N_2}=b\left(mol\right)\)
\(\Leftrightarrow a+b=1\left(1\right)\)
\(M_X=14.8\cdot2=29.6\left(\dfrac{g}{mol}\right)\)
\(m_X=29.6\cdot1=29.6\left(g\right)\)
\(\Leftrightarrow32a+28b=29.6\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.4,b=0.6\)
\(V_{O_2}:V_{N_2}=0.6:0.4=3:2\)
\(\%m_{O_2}=\dfrac{0.4\cdot32}{29.6}\cdot100\%=43.24\%\)
\(\%m_{N_2}=56.76\%\)
\(\)
