Cu không phản ứng với HCl
Mg + 2HCl \(\rightarrow\) MgCl2 + H2
=> \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
=> mMg = 0,1 . 24 =2,4 g
Lại có mCu + mMg = 10 g
=> mCu = 10 - 2,4 =7,6 g
=> %mCu = \(\dfrac{7,6}{10}.100\%=76\%\)
%mMg = 100% - 76% = 24%
PTHH:
Mg+2HCl-->MgCl2 +H2
0,1_______________0,1
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
=>\(m_{Mg}=0,1.24=2,4\left(g\right)=>\%m_{Mg}=\dfrac{2,4}{10}.100=24\%\)
=>%Cu=100-24=76%
a) Mg + 2HCl → MgCl2 + H2↑ (1)
Cu + HCl → X
b) \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
theo PT1: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,1\times24=2,4\left(g\right)\)
\(\Rightarrow\%m_{Mg}=\dfrac{2,4}{10}\times100\%=24\%\)
\(\Rightarrow\%m_{Cu}=100\%-24\%=76\%\)