Zn + 2HCl \(\rightarrow\) ZnCl2 + H2
nZn = \(\dfrac{6,5}{65}=0,1mol\)
nHCl = \(\dfrac{60.7,3\%}{36,5}=0,12mol\)
Lập tỉ lệ: nZn : nHCl = \(\dfrac{0,1}{1}:\dfrac{0,12}{2}=0,1:0,06\)
=> Zn dư
nZn dư = 0,1 - 0,06 = 0,04 mol
=> mZn dư = 0,04 . 65 = 2,6g
\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)
\(n_{HCl}=\dfrac{60\cdot7.3}{100\cdot36.5}=0.12\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(1...........2\)
\(0.1.........0.12\)
\(LTL:\dfrac{0.1}{1}>\dfrac{0.12}{2}\Rightarrow Zndư\)
\(n_{H_2}=\dfrac{0.12}{2}=0.06\left(mol\right)\)
\(V_{H_2}=0.06\cdot22.4=1.344\left(l\right)\)
\(m_{Zn\left(dư\right)}=\left(0.1-0.06\right)\cdot65=2.6\left(g\right)\)
