Câu 10:
\(m_{KOH}=\dfrac{13,44\cdot25\%}{100\%}=3,36\left(g\right)\\ \Rightarrow n_{KOH}=\dfrac{3,36}{56}=0,06\left(mol\right)\\ PTHH:3KOH+FeCl_3\rightarrow Fe\left(OH\right)_3\downarrow+3KCl\\ \Rightarrow n_{Fe\left(OH\right)_3}=\dfrac{1}{3}n_{KOH}=0,02\left(mol\right)\\ \Rightarrow m_{\downarrow}=m_{Fe\left(OH\right)_3}=0,02\cdot107=2,14\left(g\right)\\ \text{Chọn D}\)
Câu 11:
\(\%_P=\dfrac{62}{234}\cdot100\%\approx26,5\%\\ \Rightarrow m_{NTDD}=m_P\approx468\cdot26,5\%=124,02\left(g\right)\approx124\left(g\right)\\ \text{Chọn B}\)
Câu 12:
\(n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1\left(mol\right)\\ PTHH:Na_2SO_3+2HCl\rightarrow2NaCl+H_2O+SO_2\uparrow\\ \Rightarrow n_{SO_2}=0,1\left(mol\right)\\ \Rightarrow V_{SO_2}=0,1\cdot22,4=2,24\left(l\right)\\ \text{Chọn C}\)
Câu 13:
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ Fe_3O_4+8HCl\rightarrow FeCl_2+2FeCl_3+4H_2O\\ \text{Chọn B}\)
