Gọi \(\left\{{}\begin{matrix}n_{Al}=x\\n_{K_2CO_3}=y\end{matrix}\right.\) ( mol )
\(n_{hhk}=\dfrac{7,28}{22,4}=0,325\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
`x` `,15x` `0,5x` `1,5x` ( mol )
\(K_2CO_3+H_2SO_4\rightarrow K_2SO_4+CO_2\uparrow+H_2O\)
`y` `y` `y` `y` ( mol )
\(\rightarrow\left\{{}\begin{matrix}m_{muối}=342.0,5x+174y=37,92\\n_{hhk}=1,5x+y=0,325\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0,207\\y=0,0145\end{matrix}\right.\)
\(a=27.0,207+138.0,0145=7,59\left(g\right)\)
\(C\%_{H_2SO_4}=\dfrac{\left(1,5.0,207+0,0145\right).98}{200}.100=1,5925\%\)
\(m_{ddspứ}=7,59+200-\left(1,5.0,207.2\right)-\left(0,0145.44\right)=206,331\left(g\right)\)
\(\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{\left(0,5.0,207\right).342}{206,331}.100=17,15\%\\C\%_{K_2SO_4}=\dfrac{0,0145.174}{206,331}.100=1,22\%\end{matrix}\right.\)
