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If a peice of iron gains 10% of its weight due to partial rusting into Fe2O3 the percentage of total iron that has rusted is 23 13 23.3 25.67

If a peice of iron gains 10% of its weight due to partial rusting into Fe2O3 the percentage of total iron that has rusted is 
  1. 23
  2. 13
  3. 23.3
  4. 25.67

Grade:11

4 Answers

Sunil Kumar FP
askIITians Faculty 183 Points
9 years ago
4Fe + 3O2----2Fe2O3
let the weight of fe conveerted to fe2o3 be x gm
initian total weight be 100gm
now weight gain=110gm
mole of fe conveerted to fe2o3=4x/56
mole of fe2o3=x/56*2
weight of fe2o3=10x/7
now total weight is 110 gm
therefore
100-x +10x/7=110gm
x=23.3
therefore answer is c
SHARAD GUPTA
11 Points
6 years ago
4Fe+3O2---Fe2O33*32gm of O2 react with 4*56gm of FeThen 10gm react with 23.33gm If we take initial amount 100gm then it is 23.33% answer
Rehan qureshi
15 Points
5 years ago
Correct answer is 13.                                                     Mass of Fe in 4 mole =224gm and it's 10 percent =22.4gm      
Mass of fe2o3 = 171.2                                                  %of total iron that has rusted =22.4/171.2=13 .08
Saahir avroy
15 Points
5 years ago
4fe  + 3O2 - 2fe2O3
Weight of Fe2O3  is 110 gm 
Weight of Oxide must be 10 gm 
3 mole oxygen uses 4 mole fe to form rust 
So 10/32 mole will use 4/3*10/32 mole fe = 40/96 mole fe 
wt of fe reacted = 40/96 * 56 = 23.3gm 
so % of fe reacted will be 23.3% 

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