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)

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

19.2g

Explanation:

The balanced equation for the reaction is given below:

4Fe + 3O2 - > 2Fe2O3

Next, let us calculate the number of mole of O2 needed to produce 0.4 mole of Fe2O3. This is illustrated below:

From the balanced equation above,

3 moles of O2 produced 2 moles of Fe2O3.

Therefore, Xmol of O2 will produce 0.4 mole of Fe2O3 i. e

Xmol of O2 = (3x0.4) / 2

Xmol of O2 = 0.6 mole

Now let us convert 0.6 mole of O2 to grams in order to obtain the desired result. This is illustrated below:

Molar Mass of O2 = 16x2 = 32g/mol

Mole of O2 = 0.6 mol

Mass of O2 = ?

Mass = number of mole x molar Mass

Mass of O2 = 0.6 x 32

Mass of O2 = 19.2g

Therefore, 19.2g of O2 are needed to produce 0.4 mole of Fe2O3.