a) Đổi 250cm3=0,25(l)
n HCl=0,25.2=0,5(mol)
Ca+2HCl---->CaCl2+H2
x------2x-------------------x
Mg+2HCl----.MgCl2+H2
y------2y-----------------y
Theo bài ta có hpt
\(\left\{{}\begin{matrix}40x+24y=6,4\\2x+2y=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,025\\y=0,225\end{matrix}\right.\)
%m Ca=0,025.40/6,4.100%=15,625%
%m Mg=100%-15,625%=84,375%
b) CM CaCl2=0,025/0,25=0,1(M)
CM MgCl2=0,225/0,25=0,9(M)
c) n H2=1/2n HCl=0,25(mol)
V H2=0,25.22,4=5,6(l)
a) \(n_{HCl\left(pư\right)}=2.0,25.80\%=0,4\left(mol\right)\)
PTHH: Ca + 2HCl --> CaCl2 + H2
a ----> 2a ------> a ---> a (mol)
Mg +2HCl --> MgCl2 + H2
b ---> 2b -------> b -----> b (mol)
=> \(\left\{{}\begin{matrix}40a+24b=6,4\\2a+2b=0,4\end{matrix}\right.\) => \(\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,1\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}m_{Ca}=40.0,1=4\left(g\right)\\m_{Mg}=24.0,1=2,4\left(g\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%_{Ca}=\frac{4}{6,4}.100\%=62,5\%\\\%_{Mg}=\frac{2,4}{6,4}.100\%=37,5\%\end{matrix}\right.\)
b) \(Cm_{CaCl2}=\frac{0,1}{0,25}=0,4M\)
\(Cm_{MgCl2}=\frac{0,1}{0,25}=0,4M\)
c) \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
