\(n_{H2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,15 0,3 0,15 0,15
a) \(n_{Mg}=\dfrac{01,5.1}{1}=0,15\left(mol\right)\)
⇒ \(m_{Mg}=0,15.24=3,6\left(g\right)\)
b) \(n_{HCl}=\dfrac{0,15.2}{1}=0,3\left(mol\right)\)
\(m_{HCl}=0,3.36,5=10,95\left(g\right)\)
\(m_{ddHCl}=\dfrac{10,95.100}{7,3}=150\left(g\right)\)
c) \(n_{MgCl2}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)
⇒ \(m_{MgCl2}=0,15.95=14,25\left(g\right)\)
\(m_{ddspu}=3,6+150-\left(0,15.2\right)=153,8\left(g\right)\)
\(C_{MgCl2}=\dfrac{14,25.100}{153,8}=9,27\)0/0
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