\(n_{CaO}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(CaO+2HCl\rightarrow CaCl_2+H_2O\)
\(0.2...........0.4.........0.2\)
\(m_{HCl}=0.4\cdot36.5=14.6\left(g\right)\)
\(C\%_{HCl}=\dfrac{14.6}{200}\cdot100\%=7.3\%\)
\(m_{\text{dung dịch sau phản ứng}}=11.2+200=211.2\left(g\right)\)
\(m_{CaCl_2}=0.2\cdot111=22.2\left(g\right)\)
\(C\%_{CaCl_2}=\dfrac{22.2}{211.2}\cdot100\%=10.51\%\)
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