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How many grams of Fe 2 O 3 can be formed from the rusting of 446 grams of Fe according to the reaction: 4Fe + 3O 2 → 2Fe 2 O 3 and excess oxygen? 320 g 223 g 159 g 640 g

How many grams of Fe2O3 can be formed from the rusting of 446 grams of Fe according to the   reaction: 4Fe + 3O2 → 2Fe2O3 and excess oxygen?
  1. 320 g
  2. 223 g
  3. 159 g
  4. 640 g

Grade:11

1 Answers

Shivangi Khatter
askIITians Faculty 468 Points
5 years ago
dear student
the correct answer is(D)
4 *55.8 grams of Fe gives 2 160 grams of Fe2O3
446 grams of Fe = 2 *160 *446/(4 *55.8) = 640 grams

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