\(\left\{{}\begin{matrix}n_K=a\left(mol\right)\\n_{Ba}=b\left(mol\right)\end{matrix}\right.\)
\(n_{khí}=\dfrac{0,448}{22,4}=0,02mol\)
\(2K+2H_2O\rightarrow2KOH+H_2\)
\(Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\)
Từ hai pt:\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{1}{2}\\\dfrac{a}{2}+b=0,02\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,008\\b=0,016\end{matrix}\right.\)
\(\left\{{}\begin{matrix}n_{H_2SO_4}=0,05\cdot0,1=0,005\\n_{CuSO_4}=0,2\cdot0,1=0,02\end{matrix}\right.\)
\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
0,008 0,004
\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2H_2O\)
0,016 0,016 0,016
\(2KOH+CuSO_4\rightarrow K_2SO_4+Cu\left(OH\right)_2\downarrow\)
0,008 0,008
\(Ba\left(OH\right)_2+CuSO_4\rightarrow Cu\left(OH\right)_2\downarrow+BaSO_4\downarrow\)
0,016 0,016 0,016
\(m_{\downarrow}=0,016\cdot233+0,008\cdot160+0,016\cdot160+0,016\cdot233=11,296g\)
