Mg + 2HCl → MgCl2 + H2↑ (1)
Fe + 2HCl → FeCl2 + H2↑ (2)
\(n_{H_2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
a) Gọi x,y lần lượt số mol của Mg và Fe
Ta có: \(24x+56y=6\) (*)
Theo Pt1: \(n_{H_2}=n_{Mg}=x\left(mol\right)\)
Theo PT2: \(n_{H_2}=n_{Fe}=y\left(mol\right)\)
Ta có: \(x+y=0,15\) (**)
Từ (*)(**) ta có: \(\left\{{}\begin{matrix}24x+56y=6\\x+y=0,15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,075\\y=0,075\end{matrix}\right.\)
Vậy \(n_{Mg}=0,075\left(mol\right)\Rightarrow m_{Mg}=0,075\times24=1,8\left(g\right)\)
\(n_{Fe}=0,075\left(mol\right)\Rightarrow m_{Fe}=0,075\times56=4,2\left(g\right)\)
b) \(\%m_{Mg}=\frac{1,8}{6}\times100\%=30\%\)
\(\%m_{Fe}=100\%-30\%=70\%\)
c) Theo pT1: \(n_{HCl}=2n_{Mg}=2\times0,075=0,15\left(mol\right)\)
Theo PT2: \(n_{HCl}=2n_{Fe}=2\times0,075=0,15\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,15+0,15=0,3\left(mol\right)\)
\(\Rightarrow V_{ddHCl}=\frac{0,3}{1}=0,3\left(l\right)=300\left(ml\right)\)
d) Theo pT1: \(n_{MgCl_2}=n_{Mg}=0,075\left(mol\right)\)
\(\Rightarrow C_{M_{MgCl_2}}=\frac{0,075}{0,3}=0,25\left(M\right)\)
Theo PT2: \(n_{FeCl_2}=n_{Fe}=0,075\left(mol\right)\)
\(\Rightarrow C_{M_{FeCl_2}}=\frac{0,075}{0,3}=0,25\left(M\right)\)
Đặt :
nFe = x mol
nMg= y mol
mhh= 56x + 24y = 6 (g) (1)
nH2 = 3.36/22.4=0.15 mol
Fe + 2HCl --> FeCl2 + H2
x_____2x______x_____x
Mg + 2HCl --> MgCl2 + H2
y_____2y_______y_____y
nH2 = x + y = 0.15 (2)
Giải (1) và (2) :
x = y = 0.075
mFe = 4.2g
mMg = 1.8 g
%Fe = 70%
%Mg = 30%
Từ PTHH ta thấy :
nHCl = 2nH2 = 0.3 mol
VddHCl = 0.3/1=0.3 l
CM FeCl2 = CM MgCl2 = 0.075/0.3 = 0.25M
