\(X\left\{{}\begin{matrix}Cu:a\left(mol\right)\\S:b\left(mol\right)\end{matrix}\right.\underrightarrow{^{H2SO4}}\left\{{}\begin{matrix}CuSO4\\SO_2\end{matrix}\right.\)
\(n_{SO2}=0,54\left(mol\right)\)
Giải hệ PT:
\(\left\{{}\begin{matrix}64a+32b=15,36\\2a+6b=0,54.2=1,08\left(BT.e\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=0,18\\b=0,12\end{matrix}\right.\)
\(n_{CuSO2}=n_{Cu}=0,18\left(mol\right)\)
\(CuSO_4+Ba\left(OH\right)_2\rightarrow BaSO_4+Cu\left(OH\right)_2\)
0,18_______________0,18_______0,18_______
\(\Rightarrow m_{\downarrow}=m_{Cu\left(OH\right)_2}+m_{BaSO_4}=59,58\left(g\right)\)
