Question

Answer 1

\(n_{H_2}=\dfrac{0,448}{22,4}=0,02(mol)\\ 1,Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O\\ Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O\\ Fe+H_2SO_4\to FeSO_4+H_2\\ 2,n_{Fe}=0,02(mol)\Rightarrow n_{Fe_2O_3}=0,01(mol)\\ \Rightarrow m_{Fe_2O_3}=0,01.160=1,6(g)\\ \Rightarrow m_{Al_2O_3}=3,13-1,6=1,53(g)\\ 3,n_{Al_2O_3}=\dfrac{1,53}{102}=0,015(mol)\\ \Rightarrow \Sigma n_{H_2SO_4}=0,015.3+0,02=0,065(mol)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{0,065}{0,1}=0,65(l)\)