\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
a
Gọi \(\left\{{}\begin{matrix}n_K=x\left(mol\right)\\n_{Na}=y\left(mol\right)\end{matrix}\right.\)
\(K+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\)
x -------------->x------->0,5x
\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
y---------------->y---------->0,5y
Có: \(n_{H_2}=0,5x+0,5y=0,1\left(mol\right)\) (1)
\(KOH+HCl\rightarrow KCl+H_2O\)
x--------->x------->x
\(NaOH+HCl\rightarrow NaCl+H_2O\)
y---------->y--------->y
Có: \(m_{hh.muối}=74,5x+58,5y=13,3\left(g\right)\left(2\right)\)
Từ (1), (2) có hệ phương trình: \(\left\{{}\begin{matrix}0,5x+0,5y=0,1\\74,5x+58,5y=13,3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\) (bấm máy giải hệ)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Na}=\dfrac{23.0,1.100\%}{23.0,1+39.0,1}=37,1\%\\\%m_K=\dfrac{39.0,1.100\%}{23.0,1+39.0,1}=62,9\%\end{matrix}\right.\)
b
\(n_{HCl}=x+y=0,1+0,1=0,2\left(mol\right)\\ V_{HCl}=\dfrac{n}{CM}=\dfrac{0,2}{0,5}=0,4\left(l\right)\)
c
\(2H_2+O_2\underrightarrow{t^o}2H_2O\)
0,1-->0,05
\(V_{O_2}=0,05.22,4=1,12\left(l\right)\)