Bài 3:
Ta có: \(n_{CuO}=\dfrac{0,8}{80}=0,01\left(mol\right)\), \(n_{H_2SO_4}=0,3.0,5=0,15\left(mol\right)\)
a, PT: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
b, Xét tỉ lệ: \(\dfrac{0,01}{1}< \dfrac{0,15}{1}\), ta được H2SO4 dư.
Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{CuSO_4}=n_{CuO}=0,01\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,15-0,01=0,14\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,14.98=13,72\left(g\right)\)
c, \(C_{M_{CuSO_4}}=\dfrac{0,01}{0,3}\approx0,033\left(M\right)\)
\(C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0,14}{0,2}\approx0,467\left(M\right)\)
Bài 4:
Ta có: \(n_{MgO}=\dfrac{1,6}{40}=0,04\left(mol\right)\), \(n_{HCl}=\dfrac{300.7,3\%}{36,5}=0,6\left(mol\right)\)
a, \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
b, Xét tỉ lệ: \(\dfrac{0,04}{1}< \dfrac{0,6}{2}\), ta được HCl dư.
Theo PT: \(n_{HCl\left(pư\right)}=2n_{MgO}=0,08\left(mol\right)\)
\(\Rightarrow n_{HCl\left(dư\right)}=0,6-0,08=0,52\left(mol\right)\)
\(\Rightarrow m_{HCl\left(dư\right)}=0,52.36,5=18,98\left(g\right)\)
c, Ta có: m dd sau pư = mMgO + m dd HCl = 1,6 + 300 = 301,6 (g)
Theo PT: \(n_{MgCl_2}=n_{MgO}=0,04\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,04.95}{301,6}.100\%\approx1,26\%\\C\%_{HCl\left(dư\right)}=\dfrac{18,98}{301,6}.100\%\approx6,29\%\end{matrix}\right.\)
