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The formation of rust (Fe2O3) from solid iron and oxygen
Chapter 6, Problem 122CQ(choose chapter or problem)
The formation of rust (Fe2O3) from solid iron and oxygen gas releases 1.7 × 103 kJ. (6.9)4Fe(s) + 3O2(g) ? 2Fe2O3(s) ?H = 1.7 × 103 kJa. How many kilojoules are released when 2.00 g of Fe reacts?b. How many grams of rust form when 150 kcal are released?c. What is the complete equation (including heat) for the formation of rust?
Questions & Answers
QUESTION:
The formation of rust (Fe2O3) from solid iron and oxygen gas releases 1.7 × 103 kJ. (6.9)4Fe(s) + 3O2(g) ? 2Fe2O3(s) ?H = 1.7 × 103 kJa. How many kilojoules are released when 2.00 g of Fe reacts?b. How many grams of rust form when 150 kcal are released?c. What is the complete equation (including heat) for the formation of rust?
ANSWER:Solution 122CQStep 1 of 3:Given reaction:4Fe(s) + 3O2(g) 2Fe2O3(s) H = 1.7 × 103 kJa. Here, we are asked to find kilojoules, released when 2.00 g of Fe reacts.Molar mass of Fe = 55.845 g/molH = 1.7 × 103 kJ1 mol of Fe = 55.845 g of Fe. So, the mole-mass factor is: and In the reaction we see, there are 4 moles of Fe and it is equal to H = 1.7 × 103 kJ. So, the factor is: and Thus, the energy released in kilojoules is: = 2.00 g Fe = 15.22 kJHence, the kilojoules are released when 2.00 g of Fe reacts is 15.22 kJ.________________