Consider the following reaction of 5.4 g of iron with O2?

Question

4 Fe + 3 O2---> 2 Fe2O3 is the balanced formula

1)How many moles of iron react?

2)How many moles of molecular oxygen must react?

3)How many moles of Fe2O3 are formed?

4)What is the mass, in grams, of the Fe2O3?

Zaf · Accepted Answer

1) Provided that iron is the limiting reactant (5.4 g react fully),
moles of iron = 5.4g / 55.8g/mol = 0.097moles

2) Looking at the coefficients of the balanced equation, four moles of iron react with three moles of oxygen. Therefore,
moles of reactant O2 = moles of reactant iron x 3/4
= 0.097moles x 3/4 = 0.073

3) moles of Fe2O3 formed = moles of reactant iron x 2/4
= 0.097moles / 2 = 0.0485moles

4) mass = number of moles x molar mass
= 0.0485moles x 159.6g/mole = 7.7g

Hope that was clear enough. good luck with chem =)

Anonymous · Answer

1.  n =m/M
n = 5.4 / 55.86
n = 0.09667moles

2. 0.09667 / 4 x 3 = 0.0725 moles of oxygen must react

3. 0.0483moles of Fe2O3 is formed

4. n=m/M
0.0483 = m/(2x55.86+3x16)
m=7.72g

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Consider the following reaction of 5.4 g of iron with O2?

2 Answers