How many grams of O2 are needed to produce 0.400 mole Fe2O3 in the following reaction?
4 Fe(s) + 3 028 2 Fe2O3(s)

To produce 0.400 moles of Fe₂O₃, 95.8 g of O₂ are needed.

Let's consider the following balanced equation.

4 Fe(s) + 3 O₂(g) ⇒ 2 Fe₂O₃(s)

We can calculate the grams of O₂ needed to produce 0.400 moles of Fe₂O₃ considering the following relationships.

To produce 0.400 moles of Fe₂O₃, 95.8 g of O₂ are needed.

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How many grams of O₂ are needed to produce 0.400 mole Fe₂O₃ in the following reaction?

4 Fe(s) + 3 O₂(g) ⇒ 2 Fe₂O₃(s)

Answer

7.2 g of Oxygen are needed to produce 0.400 mole Fe₂O₃ in the following reaction

Explanation:

4 Fe(s) + 3O₂= 2 Fe₂O₃(s)

4 moles of Fe reacts with 3 moles of oxygen to give 2 moles of Fe₂O₃

to get the grams of oxygen needed to produce 0.400 mole Fe₂O₃ in the following reaction

we first get the moles then convert it to grams

3 moles of oxygen are needed to give 2 moles of Fe₂O₃ in the equation thus we have 3:2

let the represent the moles of oxygen that will be intended for 0.400 mole Fe₂O₃ in the following reaction

x:0.4

3:2=x:0.4

3/2 = x/0.4

cross multiply

3x0.4 = 2x

1.2 = 2x

x= 1.2/2 = 0.6

moles is 0.6

1 mole of Oxygen is 12g of Oxygen

0.6 mole of oxygen will give = 0.6 x12= 7.2 g of Oxygen

7.2 g of Oxygen are needed to produce 0.400 mole Fe₂O₃ in the following reaction

- lead + phosphoric acid = hydrogen + lead(ii) phosphate