a, Fe + 2HCl \(\rightarrow\) FeCl2 + H2
b, \(n_{Fe}=\dfrac{16,8}{56}=0,3mol\\ n_{H_2}=0,3mol\\ V_{H_2}=0,3.22,4=6,72l\)
c, \(m_{FeCl_2}=0,3.\left(56+35,5.2\right)=38,1g\)
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help me ;-;
