\(GS:n_A=1\left(mol\right)\)
\(n_{N_2}=a\left(mol\right),n_{H_2}=b\left(mol\right)\)
\(\Leftrightarrow a+b=1\left(1\right)\)
\(M_A=0.35\cdot28=9.8\left(\dfrac{g}{mol}\right)\)
\(m_A=9.8\cdot1=9.8\left(g\right)\)
\(\Leftrightarrow28a+2b=9.8\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.3,b=0.7\)
\(\%n_{N_2}=\dfrac{0.3}{1}\cdot100\%=30\%\)
\(\%n_{H_2}=70\%\)
\(\%m_{N_2}=\dfrac{0.3\cdot28}{9.6}\cdot100\%=87.5\%\)
\(\%m_{H_2}=12.5\%\)
\(\)
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a) Gọi n N2 = a(mol) ; n H2 = b(mol)
Coi n A = 1(mol)
M A = 0,35.28 = 9,8 (g/mol)
Ta có :
a + b = 1
28a + 2b = 9,8.1
Suy ra a = 0,3 ; b = 0,7
%V N2 = 0,3/1 .100% = 30%
%V H2 = 0,7/1 .100% = 70%
b)
%m N2 = 0,3.28/9,8 . 100% = 85,71%
%m H2 = 100%- 85,71% = 14,29%
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