Đáp án:
\(\begin{array}{l} a,\ V_{H_2SO_4}=0,6\ lít.\\ b,\ C_{M_{Fe_2(SO_4)_3}}=\dfrac{2}{3}\ M.\\ c,\ m_{\text{dd KOH}}=134,4\ g.\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} a,\ PTHH:Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O\\ n_{Fe}=\dfrac{6,4}{160}=0,04\ mol.\\ Theo\ pt:\ n_{H_2SO_4}=3n_{Fe_2O_3}=0,12\ mol.\\ \Rightarrow V_{H_2SO_4}=\dfrac{0,12}{2}=0,06\ lít.\\ b,\ Theo\ pt:\ n_{Fe_2(SO_4)_3}=n_{Fe_2O_3}=0,04\ mol.\\ \Rightarrow C_{M_{Fe_2(SO_4)_3}}=\dfrac{0,04}{0,06}=\dfrac{2}{3}\ lít.\\ d,\ PTHH:H_2SO_4+2KOH\to K_2SO_4+2H_2O\\ Theo\ pt:\ n_{KOH}=2n_{H_2SO_4}=0,24\ mol.\\ \Rightarrow m_{\text{dd KOH}}=\dfrac{0,24\times 56}{10\%}=134,4\ g.\end{array}\)
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