Use the following reaction to answer the questions
4Fe(s) + 3O2(g) ® 2Fe2O3(g) ÄH = -1700 kJ
How many kJ are released when 2.00 g Fe react?
How many grams of rust form when 453 kJ of energy are released?
Do these as a proportion..
2gramsFe/4molesFe=Energy/-1700kj
of course, convert the 4 moles to grams.
gramsrust/2molesFe2O3=-453/-1700
To answer the questions, we need to use stoichiometry and the given reaction equation.
Question 1: How many kJ are released when 2.00 g of Fe react?
Step 1: Convert the given mass of Fe (2.00 g) to moles.
To do this, we need to know the molar mass of Fe, which is approximately 55.845 g/mol.
Using the formula: number of moles = mass / molar mass
Number of moles of Fe = 2.00 g / 55.845 g/mol = 0.03574 mol (rounded to five decimal places)
Step 2: Calculate the amount of heat released using the stoichiometry of the reaction.
According to the balanced equation, 4 moles of Fe react to produce 1700 kJ of energy.
Using this ratio, we can set up the following proportion to find the amount of heat released when 0.03574 mol of Fe reacts.
(0.03574 mol Fe / 4 mol Fe) = (x kJ / 1700 kJ)
Cross-multiplying and solving for x, we get: x = (0.03574 mol Fe * 1700 kJ) / 4 mol Fe
x ≈ 15.273 kJ (rounded to three decimal places)
Therefore, when 2.00 g of Fe react, approximately 15.273 kJ of energy are released.
Question 2: How many grams of rust form when 453 kJ of energy are released?
Step 1: Convert the given energy (453 kJ) to moles of Fe2O3 produced.
According to the balanced equation, 1700 kJ of energy are released when 2 moles of Fe2O3 are produced.
Using this ratio, we can set up the following proportion to find the moles of Fe2O3 formed.
(1700 kJ / 2 moles Fe2O3) = (453 kJ / x moles Fe2O3)
Cross-multiplying and solving for x, we get: x = (453 kJ * 2 moles Fe2O3) / 1700 kJ
x ≈ 0.534 moles (rounded to three decimal places)
Step 2: Convert moles of Fe2O3 to grams.
To do this, we need to know the molar mass of Fe2O3, which is approximately 159.688 g/mol.
Using the formula: mass = moles * molar mass
Mass of Fe2O3 = 0.534 moles * 159.688 g/mol
Mass of Fe2O3 ≈ 85.203 g (rounded to three decimal places)
Therefore, when 453 kJ of energy are released, approximately 85.203 grams of rust (Fe2O3) are formed.