What Weight of Iron(||) sulphide is formed when 40g of sulphur reacts with Iron? (Atomic no. of Iron = 56, sulphur =32).
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Weight of S =40g
S. + Fe →FeS
32g+56g.=88g
Therefore, 32g of S react to give 88g of FeS
Then, 40g of S to give =88/32×40
= 110g of FeS
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S. + Fe →FeS
32g+56g.=88g
Therefore, 32g of S react to give 88g of FeS
Then, 40g of S to give =88/32×40
= 110g of FeS
(Ans)
Hope it Helps...!
sabika1:
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Answer:
atomic mass of Fe(iron)= 56
atomic mass of sulphur= 32
reaction: Fe + S = FeS
56 + 32= 56 + 32
56 + 32= 88g of ferrous Sulphide
32 g of sulphur react with iron to give
88 g of iron sulphide.
40g of sulphur will give = 88/32×40
= 110g of ferrous Sulphide(FeS)
Explanation:
HENCE, 40 g of sulphur will give 110 g of FeS(ferrous sulphide)...
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