If 279 grams of iron react with an excess of oxygen, as shown in the balanced equation below, how many grams of iron (iii) oxide can be formed? 4fe + 3o2 yields 2fe2o3 139.5 grams 199.5 grams 399 grams 1596 grams

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399 gram

Explanation:-

The balanced chemical equation for the reaction is

4Fe +3 O2 --> 2 Fe2O3

Atomic mass of Iron Fe is 55.845 gram

Molecular mass of Iron (III) Oxide Fe2O3 is

55.845 x 2 + 16 x 3 = 159.69 gram

From the balanced chemical equation,

4 Fe gives 2 Iron (III) oxide

∴ 4 x 55.845 grams of Fe gives 2 x 159.69 grams of Iron (III) Oxide.

279 grams of Fe gives (2 x 159.69) x 279 / ( 4 x 55.845)

= 399 gram

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.