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a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)
b, \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)
c, \(C_{M_{ZnCl_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1 0,1
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\\
V_{H_2}=0,1.22,4=2,24l\\
m_{\text{dd}}=6,5+200-\left(0,1.2\right)=206,3g\)
bài 2 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\
pthh:Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2 0,4 0,2 0,2
\(m_{HCl}=0,4.36,5=14,6g\\
V_{H_2}=0,2.22,4=4,48l\\
m\text{dd}=4,8+200-0,4=204,4g\\
C\%=\dfrac{0,2.136}{204,4}.100\%=13,3\%\)
a+b+c) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)=n_{ZnCl_2}=n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\\V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)
d) PTHH: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Theo PTHH: \(n_{H_2}=n_{Cu}=0,1\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,1\cdot64=6,4\left(g\right)\)
\(a) Zn + 2HCl \to ZnCl_2 + H_2\\ n_{ZnCl_2} = n_{Zn} = \dfrac{6,5}{65} = 0,1(mol)\\ m_{ZnCl_2} = 0,1.136 = 13,6(gam)\\ b) n_{H_2} = n_{Zn} = 0,1(mol) \Rightarrow V_{H_2} = 0,1.22,4 =2 ,24(lít)\\ c) n_{HCl} =2 n_{H_2} = 0,2(mol)\\ \Rightarrow m_{HCl} = 0,2.36,5 = 7,3(gam)\ ; V_{dd\ HCl} = \dfrac{0,2}{0,5} = 0,4(lít)\)
\(a,Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{ZnCl_2}=n_{H_2}=n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ b,m_{ZnCl_2}=136.0,1=13,6\left(g\right);V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ c,Vì:\dfrac{0,1}{1}>\dfrac{0,2}{1}\Rightarrow n_{Zn\left(TT\right)}=0,1\left(mol\right);n_{Zn\left(LT\right)}=n_{ZnCl_2}=0,2\left(mol\right)\\ H=\dfrac{0,1}{0,2}.100\%=50\%\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{H_2}=n_{Zn}=0,2\left(mol\right)\\n_{HCl}=2n_{Zn}=0,4\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(C_{M_{HCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)
$a\big)$
$n_{Zn}=\dfrac{3,25}{65}=0,05(mol)$
$Zn+2HCl\to ZnCl_2+H_2$
Theo PT: $n_{ZnCl_2}=n_{Zn}=0,05(mol)$
$\to m_{ZnCl_2}=0,05.136=6,8(g)$
$b\big)$
Theo PT: $n_{HCl}=2n_{Zn}=0,1(mol)$
$\to V_{dd\,HCl}=\dfrac{0,1}{0,5}=0,2(l)=200(ml)$
\(a)n_{Zn}=\dfrac{6,5}{65}=0,1mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{H_2}=n_{Zn}=0,1mol\\ V_{H_2}=0,1.24,79=2,479l\\ b)n_{HCl}=0,1.2=0,2mol\\ V_{ddHCl}=0,2:2=0,1l\)