Question

How many grams of CO are needed with an excess of Fe2O3 to produce 27.9 g Fe? Please show work. Fe2O3(s)+3CO(g)-->2Fe(s)+3CO2(g
**SHOW YOUR WORK**

Answer 1

Answer:

20.98 gram of CO

Explanation:

Limiting Reagent : The reactant which is present in less amount and get consumed after the reaction completes.

Molar mass : Mass of the substance present in 1 mole of the compound.

Molar mass of Fe = 55.84 gram/mole

1 mole Fe = 55.84 g

2 mole of Fe = 2 x 55.84 g = 111.68 g..........(1)

mass of C = 12

Mass of O = 16

Molar mass of CO = mass of C + mass of O

Molar mass of CO = 28 gram

1 mole of CO = 28 g

3 mole of CO = 3 x 28 = 84 g..........(2)

We need to know which is limiting reagent to calculate the amount of product or vice-versa.

In this question , it is already given that Fe2O3  is excess reagent .Hence CO must be the limiting reagent.

The balanced equation is :

$Fe_{2}O_{3}(s)+3CO(s)\rightarrow 2Fe(s)+3CO_{2}(g)$

this equation indicates ,

3 moles of CO produce = 2 mole of Fe

or in other words,

2 moles of Fe are produced from = 3 moles of CO

111.68 gram of Fe is produced from = 84 gram of CO

(from equation (1) and (2))

So ,

1 gram of Fe is produced from =

$\frac{84}{111.68}$ of CO

27.9 g of Fe is produced from=

$\frac{84}{111.68}\times 27.9$ gram of CO

= 20.98 gram of CO

= 21 gram of CO (In round figures)