PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\\ xmol:2xmol\rightarrow xmol:xmol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\\ ymol:2ymol\rightarrow ymol:ymol\)
a. Gọi x và y lần lượt là số mol của Mg và Zn.
\(m_{hh}=m_{Mg}+m_{Zn}=10,1\left(g\right)\)
\(\Leftrightarrow24x+65y=10,1\left(g\right)\left(1\right)\)
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\Leftrightarrow x+y=0,25\left(2\right)\)
Giải PT (1) và (2) ta được:
\(\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m_{Mg}=0,15.24=3,6\left(g\right)\\m_{Zn}=0,1.65=6,5\left(g\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{3,6}{\left(3,6+6,5\right)}.100\%=35,6\%\\\%m_{Zn}=100\%-35,6\%=64,4\%\end{matrix}\right.\)
b. \(m_{MgCl_2}=0,15.96=14,4\left(g\right)\)
\(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)
\(\Rightarrow m_{muoi}=14,4+13,6=28\left(g\right)\)
(?)
Mg + 2HCl → MgCl2 + H2 (1)
Zn + 2HCl → ZnCl2 + H2 (2)
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Gọi \(x,y\) lần lượt là số mol của Mg và Zn
Theo PT1: \(n_{H_2}=n_{Mg}=x\left(mol\right)\)
theo pT2: \(n_{H_2}=n_{Zn}=y\left(mol\right)\)
Ta có: \(\left\{{}\begin{matrix}24x+65y=10,1\\x+y=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\)
Vậy \(n_{Mg}=0,15\left(mol\right)\Rightarrow m_{Mg}=0,15\times24=3,6\left(g\right)\)
\(n_{Zn}=0,1\left(mol\right)\Rightarrow m_{Zn}=0,1\times65=6,5\left(g\right)\)
a) \(\%m_{Mg}=\dfrac{3,6}{10,1}\times100\%=35,64\%\)
\(\%m_{Zn}=\dfrac{6,5}{10,1}\times100\%=64,36\%\)
b) Theo PT1: \(n_{MgCl_2}=n_{Mg}=0,15\left(mol\right)\)
\(\Rightarrow m_{MgCl_2}=0,15\times95=14,25\left(g\right)\)
theo Pt2: \(n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\)
\(\Rightarrow m_{ZnCl_2}=0,1\times136=13,6\left(g\right)\)
\(\Rightarrow m_{hh}muối=14,25+13,6=27,85\left(g\right)\)
nH2 = \(\dfrac{5,6}{22,4}\) = 0,25 mol
Mg + 2HCl -> MgCl2 + H2 \(\uparrow\)
x----->2x----->x-------->x
Zn + 2HCl -> ZnCl2 + H2 \(\uparrow\)
y---->2y----->y------->y
a) ta có :
\(\left\{{}\begin{matrix}24x+65y=10,1\\x+y=0,25\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=0,15mol\\y=0,1mol\end{matrix}\right.\)
=>%Mg = \(\dfrac{0,15.24}{10,1}.100\%\) \(\approx\) 35,64 %
=>%Zn = 100% - 35,64% = 64,36%
b) mMgCl2 = 0,15 . 95 = 14,25 g
mZnCl2 = 0,1 . 136 = 13,6g
