Bài 2:
\(\%m_{Ag}=\dfrac{6.100\%}{15,3}=44,44\%\)
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
mhh Fe và Zn = 15,3-6 = 9,3 (g)
PTHH: Fe + 2HCl → FeCl2 + H2
Mol: x x
PTHH: Zn + 2HCl → ZnCl2 + H2
Mol: y y
Ta có: \(\left\{{}\begin{matrix}56x+65y=9,3\\x+y=0,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)
\(\%m_{Fe}=\dfrac{0,05.56.100\%}{15,3}=18,3\%\)
\(\%m_{Zn}=100-44,44-18,3=37,26\%\)
Bài 1:
a,\(\%m_{Ag}=\dfrac{5,4.100\%}{13,5}=40\%;\%m_{Al}=100-40=60\%\)
b,\(n_{Al}=\dfrac{13,5-5,4}{27}=0,3\left(mol\right)\)
PTHH: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Mol: 0,3 0,45
\(\Rightarrow V_{H_2}=0,45.22,4=10,08\left(l\right)\)
