a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
b) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Fe}=2.0,1=0,2\left(mol\right)\)
Đổi: 200 (ml) = 0,2 (l)
\(\Rightarrow C_{M_{HCl}}=\dfrac{n_{HCl}}{V_{HCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
c)PTHH: \(HCl+NaOH\rightarrow NaCl+H_2O\)
Theo PT: \(n_{HCl}=n_{NaOH}=0,2\left(mol\right)\)
\(\Rightarrow n_{NaOH}=0,2.40=8\left(g\right)\)
Khối lượng dung dịch NaOH 8% dùng để trung hòa lượng axit HCl 1M là:
\(m_{ddNaOH}=\dfrac{8.100}{8}=100\left(g\right)\)
nFe = \(\dfrac{5,6}{56}\)= 0,1 mol
Fe + 2HCl -> FeCl2 + H2\(\uparrow\)
0,1->0,2 mol
=> CM(HCl) = \(\dfrac{0,2}{0,2}\) = 1 M
HCl + NaOH -> NaCl + H2O
0,2->0,2 mol
=>mNaOH = 0,2 . 40 = 8 g