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Grade 11Physical Chemistry

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

Profile image of piyush soni
11 Years agoGrade 11
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4 Answers

Profile image of Sunil Kumar FP
11 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
Profile image of SHARAD GUPTA
8 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
Profile image of Rehan qureshi
8 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
Profile image of Saahir avroy
7 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%