# Sample problems corrosion

1. Question: How many grams of iron (Fe) will be corroded by 3 moles of oxygen (O₂)? Solution: The reaction is 4Fe + 3O₂ → 2Fe₂O₃. From the stoichiometry, 4 moles of Fe reacts with 3 moles of O₂. Thus, 3 moles of O₂ will react with (4/3)x3 = 4 moles of Fe. As molar mass of Fe is 55.85 g/mol, 4 moles will be 4×55.85 = 223.4 grams.

2. Question: How many moles of oxygen (O₂) are required to completely corrode 111.7 grams of iron (Fe)? Solution: The reaction is 4Fe + 3O₂ → 2Fe₂O₃. From the stoichiometry, 4 moles of Fe reacts with 3 moles of O₂. Thus, (111.7 g / 55.85 g/mol) = 2 moles of Fe will react with (3/4)x2 = 1.5 moles of O₂.
3. Question: How many grams of aluminium (Al) will be corroded by 5 moles of oxygen (O₂)? Solution: The reaction is 4Al + 3O₂ → 2Al₂O₃. From the stoichiometry, 4 moles of Al reacts with 3 moles of O₂. Thus, 5 moles of O₂ will react with (4/3)x5 = 6.67 moles of Al. As molar mass of Al is 26.98 g/mol, 6.67 moles will be 6.67×26.98 = 179.86 grams.
4. Question: How many moles of oxygen (O₂) are required to completely corrode 107.8 grams of aluminium (Al)? Solution: The reaction is 4Al + 3O₂ → 2Al₂O₃. From the stoichiometry, 4 moles of Al reacts with 3 moles of O₂. Thus, (107.8 g / 26.98 g/mol) = 4 moles of Al will react with (3/4)x4 = 3 moles of O₂.
5. Question: How many grams of copper (Cu) will be corroded by 2 moles of oxygen (O₂)? Solution: The reaction is 2Cu + O₂ → 2CuO. From the stoichiometry, 2 moles of Cu reacts with 1 mole of O₂. Thus, 2 moles of O₂ will react with 2×2 = 4 moles of Cu. As molar mass of Cu is 63.55 g/mol, 4 moles will be 4×63.55 = 254.2 grams.
6. Question: How many moles of oxygen (O₂) are required to completely corrode 127.1 grams of copper (Cu)? Solution: The reaction is 2Cu + O₂ → 2CuO. From the stoichiometry, 2 moles of Cu reacts with 1 mole of O₂. Thus, (127.1 g / 63.55 g/mol) = 2 moles of Cu will react with (1/2)x2 = 1 mole of O₂.
7. Question: How many grams of silver (Ag) will be corroded by 1 mole of sulfur (S)? Solution: The reaction is 2Ag + S → Ag₂S. From the stoichiometry, 2 moles of Ag reacts with 1 mole of S. Thus, 1 mole of S will react with 2 moles of Ag. As molar mass of Ag is 107.87 g/mol, 2 moles will be 2×107.87 = 215.74 grams.
8. Question: How many moles of sulfur (S) are required to completely corrode 215.74 grams of silver (Ag)? Solution: The reaction is 2Ag + S → Ag₂S. From the stoichiometry, 2 moles of Ag reacts with 1 mole of S. Thus, (215.74 g / 107.87 g/mol) = 2 moles of Ag will react with (1/2)x2 = 1 mole of S.
9. Question: How many grams of calcium (Ca) will be corroded by 2 moles of water (H₂O)? Solution: The reaction is Ca + 2H₂O → Ca(OH)₂ + H₂. From the stoichiometry, 1 mole of Ca reacts with 2 moles of H₂O. Thus, 2 moles of H₂O will react with 2/2 = 1 mole of Ca. As molar mass of Ca is 40.08 g/mol, 1 mole will be 40.08 grams.
10. Question: How many moles of water (H₂O) are required to completely corrode 40.08 grams of calcium (Ca)? Solution: The reaction is Ca + 2H₂O → Ca(OH)₂ + H₂. From the stoichiometry, 1 mole of Ca reacts with 2 moles of H₂O. Thus, (40.08 g / 40.08 g/mol) = 1 mole of Ca will react with 2×1 = 2 moles of H₂O.
11. Question: How many grams of sodium (Na) will be corroded by 3 moles of water (H₂O)? Solution: The reaction is 2Na + 2H₂O → 2NaOH + H₂. From the stoichiometry, 2 moles of Na reacts with 2 moles of H₂O. Thus, 3 moles of H₂O will react with (2/2)x3 = 3 moles of Na. As molar mass of Na is 22.99 g/mol, 3 moles will be 3×22.99 = 68.97 grams.
12. Question: How many moles of water (H₂O) are required to completely corrode 45.98 grams of sodium (Na)? Solution: The reaction is 2Na + 2H₂O → 2NaOH + H₂. From the stoichiometry, 2 moles of Na reacts with 2 moles of H₂O. Thus, (45.98 g / 22.99 g/mol) = 2 moles of Na will react with (2/2)x2 = 2 moles of H₂O.