1. The balanced chemical equation for the reaction of iron with oxygen to form iron oxide is:
4Fe + 3O2 → 2Fe2O3
From the equation, we can see that for every 4 moles of iron, 2 moles of iron oxide are produced. This means that the molar ratio between iron and iron oxide is 4:2, or simplified, 2:1.
To find out how much rust (iron oxide) should be produced when 100g of iron reacts, we need to calculate the molar mass of iron and iron oxide.
The molar mass of iron (Fe) is 55.85 g/mol, and the molar mass of iron oxide (Fe2O3) is 159.69 g/mol (2*55.85 + 3*16).
Using the molar ratio of 2:1, the theoretical yield of iron oxide can be calculated as follows:
(100g Fe) * (1 mol Fe / 55.85 g Fe) * (1 mol Fe2O3 / 2 mol Fe) * (159.69 g Fe2O3 / 1 mol Fe2O3) = 142.53 g Fe2O3
Therefore, 142.53 grams of rust (iron oxide) should be produced.
2. The percentage yield can be calculated by dividing the actual yield (130g) by the theoretical yield (142.53g) and multiplying by 100.
Percentage yield = (130g / 142.53g) * 100 = 91.10%
Thus, the percentage yield is approximately 91.10%.
iron reacts with oxygen to form iron oxide.
1. if 100g of iron reacts with excess oxygen, how much rust should be produced?
2 if the actual mass of iron oxide at the end of the experiment is130g, what is the percentage yield?
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