Iron is extracted from fe 2o 3 by a reaction with carbon (c) at very high temperature. if you take 2.86 x 10 4 g of fe 2o 3 to react with 9.05 x 10 3 g of c, how many moles of fe 2o 3 will be used? 2fe 2o 3 + 3c → 4fe + 3co 2

Balance Chemical Equation is as follow,

2 Fe₂O₃ + 3 C → 4Fe + 3 CO₂

According to this balance equation 319.38 g (2 moles) of Fe₂O₃ completely converts into product when reacted with 36 g (3 mole) of C.

Calculating amount of C required to consume 2.86 * 10⁴ g of Fe₂O₃,

As,

319.38 g Fe₂O₃ was consumed by = 36 g of C

Then

2.86 * 10⁴ g of Fe₂O₃ will be consumed by = X g of C

Solving for X,

X = (36 g * 2.86 * 10⁴ g) : 319.38 g

X = 3.22 * 10³ g of C.

So,

3.22 * 10³ g of C is required to consume 2.86 * 10⁴ g of Fe₂O₃, But, we are provided with 9.05 * 10³ g of C, means C is in excess and Fe₂O₃ is Limiting Reagent and will control the amount of product formed.

So,

As,

319.38 g of Fe₂O₃ produced = 223.36 g of Fe

Then

2.86 * 10⁴ g of Fe₂O₃ will produce = X g of Fe

Solving for X,

X = (2.86 * 10⁴ g * 223.36 g) : 319.38 g

X = 20001 g

Or,

X = 2.0 * 10⁴ g of Fe

Balanced equation:

2 Fe₂O₃ + 3 C → 4 Fe + 3 CO₂

number of moles of Fe₂O₃ = weight (g) / molar mass = (2.86 x 10⁴ g) / 159.69 g/mol = 179 moles

number of moles of C = (9.05 x 10³) / 12 = 754.16 moles

2 moles of Fe₂O₃ reacts with 3 moles of C

so 179 moles of Fe₂O₃ need 268.5 moles of C only

so carbon present in excess (754.16) and Fe₂O₃ is the limiting reactant so the number of moles used from Fe₂O₃ will be all the 179 moles