\(m_{\downarrow}=m_{Cu\left(OH\right)_2}+m_{BaSO_4}=98b+0.18\cdot233=59.58\left(g\right)\)
\(\Rightarrow b=0.18\)
Bảo toàn điện tích :
\(a=\dfrac{0.18\cdot2+0.24-0.18\cdot2}{3}=0.08\left(mol\right)\)
\(m_M=0.08\cdot27+0.18\cdot64+0.24\cdot35.5+0.18\cdot96=39.48\left(g\right)\)
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