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?
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%.
To answer these questions, we need to use stoichiometry, which is a method used in chemistry to calculate the amounts of reactants and products in a chemical reaction.
1. The balanced equation for the reaction between iron and oxygen is:
4Fe + 3O2 -> 2Fe2O3
From the balanced equation, we can see that 4 moles of iron react with 3 moles of oxygen to produce 2 moles of iron oxide.
To find the amount of iron oxide produced, we need to calculate the number of moles of iron, and then use the stoichiometric ratio to determine the moles of iron oxide.
The molar mass of iron (Fe) is 55.845 g/mol.
So, the number of moles of iron can be calculated as:
moles of iron = mass of iron / molar mass of iron
moles of iron = 100g / 55.845 g/mol ≈ 1.787 mol
Using the stoichiometric ratio, we can find the moles of iron oxide produced:
moles of iron oxide = (moles of iron / 4) × 2
moles of iron oxide = (1.787 mol / 4) × 2 ≈ 0.894 mol
Now, we can calculate the mass of rust (iron oxide) produced:
mass of rust = moles of iron oxide × molar mass of iron oxide
mass of rust = 0.894 mol × (2 × 55.845 g/mol) ≈ 99.998 g
Therefore, approximately 99.998 g of rust (iron oxide) should be produced.
2. The theoretical yield of iron oxide is calculated based on the stoichiometry of the reaction. In this case, the theoretical yield would be the mass of rust (iron oxide) calculated in question 1, which is approximately 99.998 g.
To calculate the percentage yield, we need to compare this theoretical yield to the actual mass of iron oxide obtained in the experiment.
Percentage yield = (Actual mass / Theoretical mass) × 100
Percentage yield = (130g / 99.998g) × 100 ≈ 130.001%
Therefore, the percentage yield is approximately 130.001%.