\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{HCl}=\dfrac{100.14,6\%}{36,5}=0,4\left(mol\right)\)
Pt: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1mol 0,4mol \(\rightarrow\) 0,1mol
Lập tỉ số: \(n_{Fe}:n_{HCl}=0,1< 0,4\)
\(\Rightarrow\)Fe hết, HCl dư
\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b) Có
\(n_{H_2SO_4\left(dư\right)}=0,4-0,1=0,3\left(mol\right)\)
Pt: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
0,6mol \(\leftarrow\) 0,3mol
\(V_{NaOH}=\dfrac{0,6}{1}=0,6\left(l\right)=600\left(ml\right)\)