a) Ta có: \(m_{dd.tăng}=m_{KL}-m_{H_2}\Rightarrow m_{H_2}=11-10,2=0,8\left(g\right)\)
`=>` \(n_{H_2}=\dfrac{0,8}{2}=0,4\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\Rightarrow56x+27y=11\left(1\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
x---------------------------->x
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
y------------------------------->1,5y
`=> x + 1,5y = 0,4 (2)`
Từ `(1), (2) => x = 0,1; y = 0,2`
`=>` \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{11}.100\%=50,9\%\\\%m_{Al}=100\%-50,9\%=49,1\%\end{matrix}\right.\)
b) \(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH: \(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\)
0,4<--------------------0,6
Áp dụng ĐLBTKL: \(m_{KClO_3\left(bđ\right)}=0,6.32+42,05=61,25\left(g\right)\)
`=>` \(n_{KClO_3\left(bđ\right)}=\dfrac{61,25}{122,5}=0,5\left(mol\right)\Rightarrow H=\dfrac{0,4}{0,5}.100\%=80\%\)
c) Ta có: \(\left\{{}\begin{matrix}m_C=m_X+m_Y=m_{H_2}+m_{O_2}=0,8+32.0,6=20\left(g\right)\\n_C=n_X+n_Y=n_{H_2}+n_{O_2}=0,4+0,6=1\left(mol\right)\end{matrix}\right.\)
`=>` \(M_C=\dfrac{20}{1}=20\left(g/mol\right)\)
`=>` \(d_{C/He}=\dfrac{20}{4}=5\)
