Question

How many grams of CO are needed to react with an excess of fe2o3 to produce 209.7 g fe

Answer 1

There are 157.8 grams of CO needed.

Step-by-step explanation:

Step 1: Data given

Mass of Fe produced = 209.7 grams

Molar mass of Fe2O3 = 159.69 g/mol

Molar mass of Fe = 55.845 g/mol

Step 2: The balanced equation

3CO + Fe2O3 → 2Fe + 3CO2

Step 3: Calculate Moles of Fe

Moles Fe = mass Fe / molar mass Fe

Moles Fe = 209.7 grams / 55.845 g/mol

Moles Fe = 3.755 moles

Step 4: Calculate moles of CO

For 3 moles of CO we need 1 mol of Fe2O3 to produce 2 moles Fe and 3 moles of CO2

For 3.755 moles of Fe we need 3.755 *3/2 = 5.6325 moles of CO

Step 5: Calculate mass of CO

Mass CO = moles CO * molar mass CO

Mass CO = 5.6325 * 28.01 g/mol

Mass CO = 157.8 grams

There are 157.8 grams of CO needed.

Answer 2

Amount of CO required is 157.5 g

Step-by-step explanation:

Molecular mass of Fe = 55.845 g/mol

Amount of Fe = 209.7 g

`Mol\ of\ Fe=(209.7)/(55.845) \\=3.75\ mol`

Balanced reaction of reduction of Fe2O3 is as follows:

`Fe_2O_3(s) + 3CO (g)\rightarrow 2Fe(s)+3CO_2(g)`

From the balanced reaction, 2 mol of Fe is produced by 3 mol of CO.

Therefore, 3.75 mol of Fe will be produced by,

`(3)/(2) * 3.75 = 5.625\ mol\ CO`

Relation between mass and mol is as follows:

Mass = Mol × Molecular formula

Molecular mass of CO = 28 g/mol

Grams of CO required = 5.625 mol × 28 g/mol

= 157.5 g

Amount of CO required is 157.5 g.