- \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
\(m_{HCl}=\dfrac{250.7,3}{100}=18,25\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
Lập tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\)
=> HCl dư. Mg hết => tính theo \(n_{Mg}\)
a.- Theo PT ta có: \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
- \(n_{HCl\left(pư\right)}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
\(n_{HCl\left(dư\right)}=0,5-0,4=0,1\left(mol\right)\)
\(\Rightarrow m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)
b. Dung dịch sau phản ứng gồm: \(MgCl_2;HCl\left(dư\right)\)
\(n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow m_{MgCl_2}=0,2.95=19\left(g\right)\)
- \(m_{H_2}=0,2.2=0,4\left(mol\right)\)
\(\Rightarrow mdd_{saupư}=4,8+250-0,4=254,4\left(g\right)\)
\(\Rightarrow\%C_{MgCl2}=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{19}{254,4}.100\%=7,46\%\)
\(\%C_{HCl\left(dư\right)}=\dfrac{3,65}{254,4}.100\%=1,43\%\)
