Bài 22:
a) \(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\); \(n_{H_2SO_4}=\dfrac{294.10\%}{98}=0,3\left(mol\right)\)
b)
PTHH: \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\) => MgO hết, H2SO4 dư
PTHH: \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
0,2------>0,2-------->0,2
=> \(m_{MgSO_4}=0,2.120=24\left(g\right)\)
\(m_{H_2SO_4\left(dư\right)}=\left(0,3-0,2\right).98=9,8\left(g\right)\)
c) mdd sau pư = 8 + 294 = 302 (g)
d) \(\left\{{}\begin{matrix}C\%_{MgSO_4}=\dfrac{24}{302}.100\%=7,95\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{9,8}{302}.100\%=3,25\%\end{matrix}\right.\)
Bài 25:
a) \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\); \(n_{HCl}=\dfrac{109,5.10\%}{36,5}=0,3\left(mol\right)\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\) => Mg hết, HCl dư
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,1---->0,2----->0,1---->0,1
=> \(m_{H_2}=0,1.2=0,2\left(g\right)\)
b) mdd sau pư = 2,4 + 109,5 - 0,2 = 111,7 (g)
c) \(\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,1.95}{111,7}.100\%=8,5\%\\C\%_{HCl\left(dư\right)}=\dfrac{\left(0,3-0,2\right).36,5}{111,7}.100\%=3,27\%\end{matrix}\right.\)
