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a sample of an oxide of iron was reduced to iron by heating with hydrogen. the mass of iron obtained was 4.35g and mass of water is 1.86g. deduce the equation for the reaction occurred.​

Respuesta :

Answer:

FeO(s) + H2(g)→ Fe(s) + H2O(g)

Explanation:

Moles of Iron will be = 4.35 g/55.845 g/mol

                                  = 0.0786 moles

Moles of water = 1.86 g/18 g/mol

                         = 0.103 moles

The mole ratio of Iron to water

= 0.0786 : 0.103

= 1   :  1.3

= 1 : 1

Therefore, the equation for the reaction is;

FeO(s) + H2(g)→ Fe(s) + H2O(g)

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