n Zn = 0.2 mol
n HCl = 0.5 mol
a) Zn +2 HCl ----> ZnCl2 + H2
0.2 0.4 0.2 0.2
VH2 = 0.2*22.4 = 4.48 l
b) mZnCl2 = 0.2*136= 27.2g
mHCl = 0.1*36.5 = 3.65g
Zn +2HCl -----> ZnCl2 +H2
a) Ta có
n\(_{Zn}=\frac{13}{65}=0,2\left(mol\right)\)
n\(_{HCl}=0,25.2=0,5\left(mol\right)\)
=> HCl dư
Theo pthh
n\(_{H2}=n_{Zn}=0,2\left(mol\right)\)
a=V\(_{H2}=0,2.22,4=4,48\left(l\right)\)
b) Chất sau pư gồm: HCl dư, ZnCl2 và H2
Theo pthh
n\(_{HCl}=2n_{Zn}=0,4mol\)
n\(_{HCldu}=0,5-0,4=0,1mol\)
m\(_{HCldu}=0,1.36,5=3,65\left(g\right)\)
m\(_{ZnCl2}=0,2.136=27,2\left(g\right)\)
m\(_{H2}=0,2.2=0,4\left(g\right)\)
Chúc bạn học tốt
Nhớ tích cho mình nhé
\(n_{Zn}=\frac{13}{65}=0,2\left(mol\right);n_{HCl}=0,25.2=0,5\left(mol\right)\)
\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(TL:\frac{0,2}{1}< \frac{0,5}{2}\) → HCl dư
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(\left\{{}\begin{matrix}m_{H_2}=0,2.2=0,4\left(g\right)\\m_{ZnCl_2}=0,2.136=27,2\left(g\right)\\m_{HCl\cdot du}=36,5.\left(0,5-0,4\right)=3,65\left(g\right)\end{matrix}\right.\)
