n Fe = a(mol) ; n Mg = b(mol)
=> 56a + 24b = 8(1)
$Mg + 2HCl \to MgCl_2 + H_2$
$Fe + 2HCl \to FeCl_2 + H_2$
n HCl = 2a + 2b = 14,6/36,5 =0,4(2)
Từ (1)(2) => a = b = 0,1
Vậy :
m Fe = 0,1.56 = 5,6(gam)
m Mg = 0,1.24 = 2,4(gam)
n H2 = a + b = 0,2(mol)
V = 0,2.22,4 = 4,48 lít
\(n_{Fe}=a\left(mol\right),n_{Mg}=b\left(mol\right)\)
\(n_{HCl}=\dfrac{14.6}{36.5}=0.4\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
mhh= 56a + 24b = 8 (g)
nHCl = 2a + 2b = 0.4 (mol)
=>a = 0.1
b = 0.1
mFe= 0.1 * 56 = 5.6 (g)
mMg = 2.4 (g)
nH2 = nHCl/2 = 0.4/2 = 0.2 (mol)
VH2 = 0.2*22.4 = 4.48 (l)
Gọi \(n_{Fe}=x\left(mol\right);n_{Mg}=y\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\)
\(n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
Theo PTHH: \(n_{KL}=x+y=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\\ \Rightarrow x+y=0,2\)
Lại có \(m_{KL}=m_{Fe}+m_{Mg}=56x+24y=8\left(g\right)\)
Ta lập hệ pt: \(\left\{{}\begin{matrix}56x+24y=8\\x+y=0,2\end{matrix}\right.\)
Giải hệ ta được \(x=y=0,1\left(mol\right)\)
\(m_{Fe}=56.0,1=5,6\left(g\right)\\ m_{Mg}=24.0,1=2,4\left(g\right)\)
Từ PTHH có \(n_{KL}=n_{H_2}=0,2\left(mol\right)\)
\(V_{H_2\left(ĐKTC\right)}=0,2.22,4=4,48\left(l\right)\)
