Fe2O3 + 3CO -> 2Fe +3CO2
要產生2.1g的鐵 需用多少質量的氧化鐵 ( III )
O=16.0 Fe=56.0
又有化學題要問..
2008-04-05 4:01 am
回答 (2)
2008-04-05 4:34 am
no.of mol of Fe2O3 / no.of mol of Fe formed = 1/2.
no.of mol of Fe required to be formed= (2.1/56.0)g=0.0375 mol.
i.e. no. of mol of Fe2O3 required = 0.0375(1/2) mol= 0.01875 mol.
mass of Fe2O3 required = (0.01875) (160.0)g = 3.0g.
no.of mol of Fe required to be formed= (2.1/56.0)g=0.0375 mol.
i.e. no. of mol of Fe2O3 required = 0.0375(1/2) mol= 0.01875 mol.
mass of Fe2O3 required = (0.01875) (160.0)g = 3.0g.
參考: me
2008-04-05 4:11 am
the number of mole of iron formed=2.1/56
since the molar ratio of iron iii oxide to iron=1:2
the number of mole of iron iii oxide needed=2.1/56x1/2
the mass of iron iii oxide needed=2.1/56x1/2x(56x2+16x3)
=3g
since the molar ratio of iron iii oxide to iron=1:2
the number of mole of iron iii oxide needed=2.1/56x1/2
the mass of iron iii oxide needed=2.1/56x1/2x(56x2+16x3)
=3g
收錄日期: 2021-04-29 20:25:39
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