Câu 3 :
\(n_{MgO}=\dfrac{6}{40}=0,15\left(mol\right)\)
Pt : \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)
1 2 1 1
0,15 0,3 0,15
a) \(n_{MgCl2}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)
⇒ \(m_{MgCl2}=0,15.95=14,25\left(g\right)\)
\(n_{HCl}=\dfrac{0,15.2}{1}=0,3\left(mol\right)\)
\(m_{HCl}=0,3.36,5=10,95\left(g\right)\)
\(m_{ddHCl}=\dfrac{10,95.100}{10}=109,5\left(g\right)\)
\(m_{ddspu}=6+109,5=115,5\left(g\right)\)
\(C_{MgCl2}=\dfrac{14,25.100}{115,5}=12,34\)0/0
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