Fe+ 2HCl-> FeCl\(_2\)+ H\(_2\)\(\uparrow\)
1 2 1 1 (mol)
0,6 (mol)
b) m\(_{HCl}\)= \(\dfrac{C\%.mdd}{100\%}\)= \(\dfrac{10,95.200}{100}\)=21,9 (g)
n\(_{HCl}\)= \(\dfrac{m}{M}\)=\(\dfrac{21,9}{36,5}\)= 0,6 mol
n\(_{H_2}\)= 0,3 mol
V\(_{H_2}\)= n.22,4 = 0,3.22,4 = 6,72 (l)
c) n\(_{Fe}\)=0,3 mol
m\(_{Fe}\)= n.M= 0,3.56= 16,8 g
Vậy: A=16,8
d) n\(_{FeCl_2}\)= 0,3 mol
m\(_{FeCl_2}\)= n.M= 0,3. 127= 38,1(g)
m\(_{dd}\)sau phản ứng= m\(_{Fe}\)+ m\(_{dd}HCl\)- m\(_{H_2}\)= 16,8+200- (0,3.2) = 216,2(g)
C%\(_{FeCl_2}\)= \(\dfrac{m_{ct}}{m_{dd}}\).100% = \(\dfrac{38,1}{216,2}\).100= 17,62% (xấp xỉ)