\(n_{H2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
0,25 <------------------------ 0,25 (mol)
=> \(m_{Mg}=0,25.24=6\left(g\right)\)
=> \(m_{Cu}=20-6=14\left(g\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
Ta có :
\(n_{H2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow n_{Mg}=n_{H2}=0,25\left(mol\right)\)
\(\Rightarrow m_{Cu}=20-0,25.24=14\left(g\right)\)
\(\Rightarrow\%m_{Cu}=\frac{14}{20}.100\%=70\%\)
nMg=nH2=0,25 mol
mCu=20-0,25.24=14g
