\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(1..........2\)
\(0.2........0.3\)
\(LTL:\dfrac{0.2}{1}>\dfrac{0.3}{2}\) \(\Rightarrow Zndư\)
\(m_{Zn\left(dư\right)}=\left(0.2-0.15\right)\cdot65=3.25\left(g\right)\)
\(n_{ZnCl_2}=\dfrac{n_{HCl}}{2}=\dfrac{0.3}{2}=0.15\left(mol\right)\)
\(\Rightarrow m_{ZnCl_2}=0.15\cdot136=20.4\left(g\right)\)
\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
a, Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,3}{2}\), ta được Zn dư.
Theo PT: \(n_{Zn\left(pư\right)}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)
\(\Rightarrow n_{Zn\left(dư\right)}=0,05\left(mol\right)\)
\(\Rightarrow m_{Zn\left(dư\right)}=0,05.65=3,25\left(g\right)\)
b, Theo PT: \(n_{ZnCl_2}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)
\(\Rightarrow m_{ZnCl_2}=0,15.136=20,4\left(g\right)\)
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)
Bạn tham khảo nhé!
PTHH: Zn + 2HCl ==> ZnCl2 + H2 (1)
ADCT n= m/M
nzn=13/56=0,2 (mol)
Theo pt (1) có
0,2/1 < 0,3/2
-> Zn dư, HCL hết
ADCT n=m/M
nZnCl2 = 0,3/2 =0,15(mol)
Ct m=n x M
=> mZnCl2 = 0,15 x 136 = 20,4 (gam)
