Question

Fe2O3+2Al=Al2O3+2Fe A welder has 1.873 × 10^2 g Fe2O3 and 94.51 g Al in his welding kit. Which reactant will he run out of first? (options: Fe2O3 or Al.......it's Fe2O3). How much of this reactant should he order to make sure he runs out of both reactants at the same time? He should order ___g of the limiting reactant.

Answer 1

92.41g

Step-by-step explanation:

Answer 2

Answer: a)
`Fe_2O_3`

b) 92.1 g

Step-by-step explanation:

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}`

a) moles of
`Fe_2O_3`

`\text{Number of moles}=(1.873* 10^2 g)/(159.69g/mol)=1.173moles`

b) moles of
`Al`

`\text{Number of moles}=(94.51g)/(27g/mol)=3.500moles`

`Fe_2O_3+2Al\rightarrow Al_2O_3+2Fe`

According to stoichiometry :

1 mole of
`Fe_2O_3` require 2 moles of
`Al`

Thus 1.173
`Cl_2` require=
`(2)/(1)* 1.173=2.346moles` of
`Al`

Thus
`Fe_2O_3` is the limiting reagent as it limits the formation of product. Thus
`Fe_2O_3` will run out first.

`Al` is the excess reagent as (3.500-2.346)= 1.154 moles are left unreacted.

2 moles of
`Al` require 1 mole of
`Fe_2O_3`

3.500 moles of
`Al` require=
`(1)/(2)* 3.500=1.750moles` of
`Fe_2O_3`

Moles of
`Fe_2O_3` required = (1.750-1.173) = 0.577

Mass of
`Fe_2O_3=moles* {\text {Molar mass}}=0.577moles* 159.69g/mol=92.1g`

Thus he should order 92.1 g of the limiting reactant.