\(n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.15.......0.3.................0.15\)
\(m_{Fe}=0.15\cdot56=8.4\left(g\right)\)
\(m_{dd_{HCl}}=\dfrac{0.3\cdot36.5}{20\%}=54.75\left(g\right)\)
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