A) 4 g
B) 32 g
C) 44 g
D) Nil
Correct Answer: C
Solution :
: \[\underset{56}{\mathop{Fe}}\,+\underset{32}{\mathop{S}}\,\to FeS\] 32g of S and 56g of Fe is required to prepare one mole of \[FeS\]. \[\therefore \]\[100-56=44g\]of iron powder will remain unreacted.You need to login to perform this action.
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