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ID: 1031286 • Letter: D
Question
develop a less costly waj tu develop an alternative to sulfuric acid that does not s You gases t does not produce toxic 9B Section Review own bel Ow 4Fe (s) +30, (g) 2Fe,O, (s) a. How many moles of rust can be produced from of iron metal? process? rust? with 50.0 g iron metal? b. How many moles of oxygen will be consumed in ts n this c. What mass of iron metal is needed to form 1000 g of d. How many molecules of oxygen will be needed to readt O 2. In regard to the reaction introduced in Question 1: a. If 200.0 g of Fe react fully with O2, what is the theoretical yield of Fe,O,? b. If the actual yield is 252.7 g, what is the percent yield?Explanation / Answer
Balanced equation
4Fe(s) + 3O2(g) ----------> 2Fe2O2(s)
4 mole 3 mole 2 mole
Molar mass (g/mole) 55.845 32.0 159.69
a.) From balanced equation we know that 4 mole of Fe yields 2 mole of rust which is half of the mole of Fe. So, the of rust produced by 3.2 mole of Fe is 3.2/2 = 1.6 mole.
b.) It is clear from balanced equation 4 mole of Fe will reacts with 3 mole of oxygen, so, 3 mole of oxygen will be consumed in the process of rusting.
c.) First we need to calculate the mole of rust
mole of rust = mass of rust/molar mass of rust = 100 g/159.69 g/mole = 0.6262 mole
From balanced equation we know that the mole of rust produced is half of the mole of Fe that means double mole of Fe will be consumed. So, the mole of Fe is 0.6262 × 2 = 1.253 mole.
Convert the mole into gram to get the mass of Fe
Mass of Fe = mole of Fe × Molar mass of Fe = 1.253 mole × 55.845 g/mol = 69.973 g
d.) First we need to calculate the mole of Fe
mole of Fe = Mass of Fe/molar mass of Fe = 50.0 g/55.845 g/mole = 0.895 mole.
From balanced equation we know that 4 mole of Fe will reacts with 3 mole of oxygen. So, 0.895 mole of Fe will reacts with 0.895 × 3/4 = 0.671 mole of oxygen.
Convert the mole into number of molecules
number of molecules = 0.671 mole × 6.023 × 1023 number of molecules/mole = 4.042 × 1023 molecules
2.
a.) Theoretical yield = Molar mass of rust/molar mass of Fe × Mass of Fe in g = 159.69/55.845 × 200 g = 571.90 g
b.) Percent yield = actual yield /Theoretical yield × 100 = 252.7 g / 571.9 × 100 = 44.186%
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