Question

Answer 1

\(n_{H2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

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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