No of moles of fe2so3 formed by combustion of 5.6 gm of fe is
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Balanced chemical equation for given reaction is 4Fe + 3O2 ⇒ 2Fe2O3
5.6gm Fe equal to moles=mass of Fe/molar mass of Fe=5.6/56=0.1 moles
From the above equation we can see that
4 moles of Fe give 2 moles of Fe2SO3
2 moles of Fe will give 1 mole of Fe2SO3
0.1 mole of Fe will give 0.05 mole of Fe2SO3 which is equal to 9.6 grams.
5.6gm Fe equal to moles=mass of Fe/molar mass of Fe=5.6/56=0.1 moles
From the above equation we can see that
4 moles of Fe give 2 moles of Fe2SO3
2 moles of Fe will give 1 mole of Fe2SO3
0.1 mole of Fe will give 0.05 mole of Fe2SO3 which is equal to 9.6 grams.
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