Bài 7:
Ta có: \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
___0,3____0,6____0,3_____0,3 (mol)
a, \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b, \(m_{HCl}=0,6.36,5=21,9\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{21,9}{21,9\%}=100\left(g\right)\)
c, Có: m dd sau pư = 16,8 + 100 - 0,3.2 = 116,2 (g)
\(\Rightarrow C\%_{FeCl_2}=\dfrac{0,3.127}{116,2}.100\%\approx32,79\%\)
Bạn tham khảo nhé!
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