Câu 9:
Ta có: \(m_{BaCl_2}=200.26\%=52\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{52}{208}=0,25\left(mol\right)\)
\(n_{H_2SO_4}=0,3.1=0,3\left(mol\right)\)
PT: \(BaCl_2+H_2SO_4\rightarrow2HCl+BaSO_{4\downarrow}\)
Xét tỉ lệ: \(\dfrac{0,25}{1}< \dfrac{0,3}{1}\), ta được H2SO4 dư.
Theo PT: \(\left\{{}\begin{matrix}n_{HCl}=2n_{BaCl_2}=0,5\left(mol\right)\\n_{BaSO_4}=n_{H_2SO_4\left(pư\right)}=n_{BaCl_2}=0,25\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,05\left(mol\right)\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)
\(m_{BaSO_4}=0,25.233=58,25\left(g\right)\)
\(m_{HCl}=0,5.36,5=18,25\left(g\right)\)
Câu 10:
\(n_{NaOH}=\dfrac{40}{40}=1\left(mol\right)\)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{1}{0,234}\approx4,2735\left(M\right)\)
Câu 11:
\(n_{HCl\left(2M\right)}=0,7.2=1,4\left(mol\right)\)
\(n_{HCl\left(3M\right)}=1,1.3=3,3\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{1,4+3,3}{0,7+1,1}\approx2,611\)
