Ta có: \(n_{H_2}=0,1\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
___0,1_____0,2__________0,1 (mol)
\(m_{Fe}=0,1.56=5,6\left(g\right)\)
\(C_{M_{HCl}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)
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