\(n_{Mg}=0,3\left(mol\right)\)
\(n_{HCl}=0,3\left(mol\right)\)
\(PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\)
...............0,15......0,3..........0,15.....0,15......
- Thấy sau phản ứng HCl phản ứng hết, Mg còn dư ( dư 0,15 mol )
\(\Rightarrow\left\{{}\begin{matrix}m_M=m_{MgCl_2}=14,25\left(g\right)\\V=V_{H_2}=3,36\left(l\right)\end{matrix}\right.\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
Ta có: \(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
\(n_{HCl}=0,2.1,5=0,3\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,3}{2}\), ta được Mg dư.
Theo PT: \(n_{MgCl_2}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,15.95=14,25\left(g\right)\\V_{H_2}=0,15.22,4=3,36\left(l\right)\end{matrix}\right.\)
Bạn tham khảo nhé!