\(1.Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{H_2}=\dfrac{0,336}{22,4}=0,015\left(mol\right)\\ TheoPT:n_{HCl}=2n_{H_2}=0,03\left(mol\right)\\ \Rightarrow V_{HCl}=\dfrac{0,03}{0,75}=0,04\left(l\right)\)
\(2.TheoPT:n_{Mg}=n_{H_2}=0,015\left(mol\right)\\ \Rightarrow a=m_{Mg}=0,015.24=0,36\left(g\right)\)
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