Question

[Đỗ Anh Thư]

dùng 6,72 lít khí hidro ở đktc để khử lần lượt là HgO, PbO, FeO, Fe₂O₃,CuO,Fe₃O₄. a)Tính khối lượng kim loại được tạo ra. b)Tính khối lượng mỗi oxit cần dùng

Answer 1

[๖ۣۜDũ๖ۣۜN๖ۣۜG]

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: \(HgO+H_2\underrightarrow{t^o}Hg+H_2O\) (1)

            \(PbO+H_2\underrightarrow{t^o}PbO+H_2O\) (2)

            \(FeO+H_2\underrightarrow{t^o}Fe+H_2O\) (3)

            \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\) (4)

            \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\) (5) 

            \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\) (6)

a) \(n_{Hg\left(1\right)}=0,3\left(mol\right)\Rightarrow m_{Hg\left(1\right)}=0,3.201=60,3\left(g\right)\)

    \(n_{Pb\left(2\right)}=0,3\left(mol\right)\Rightarrow m_{Pb\left(2\right)}=0,3.207=62,1\left(g\right)\)

   \(n_{Fe\left(3\right)}=0,3\left(mol\right)\Rightarrow m_{Fe\left(3\right)}=0,3.56=16,8\left(g\right)\)

   \(n_{Fe\left(4\right)}=0,2\left(mol\right)\Rightarrow m_{Fe\left(4\right)}=0,2.56=11,2\left(g\right)\)

   \(n_{Cu\left(5\right)}=0,3\left(mol\right)\Rightarrow m_{Cu\left(5\right)}=0,3.64=19,2\left(g\right)\)

   \(n_{Fe\left(6\right)}=0,225\left(mol\right)\Rightarrow m_{Fe\left(6\right)}=0,225.56=12,6\left(g\right)\)

b)

\(n_{HgO\left(1\right)}=0,3\left(mol\right)\Rightarrow m_{HgO\left(1\right)}=0,3.217=65,1\left(g\right)\)

\(n_{PbO\left(2\right)}=0,3\left(mol\right)\Rightarrow m_{PbO\left(2\right)}=0,3.223=66,9\left(g\right)\)

\(n_{FeO\left(3\right)}=0,3\left(mol\right)\Rightarrow m_{FeO\left(3\right)}=0,3.72=21,6\left(g\right)\)

\(n_{Fe_2O_3\left(4\right)}=0,1\left(mol\right)\Rightarrow m_{Fe_2O_3\left(4\right)}=0,1.160=16\left(g\right)\)

\(n_{CuO\left(5\right)}=0,3\left(mol\right)\Rightarrow m_{CuO\left(5\right)}=0,3.80=24\left(g\right)\)

\(n_{Fe_3O_4\left(6\right)}=0,075\left(mol\right)\Rightarrow m_{Fe_3O_4}=0,075.232=17,4\left(g\right)\)