Question

For the following reaction if you have 13.2g of CO and 42.7g of Fe2O3, which is the limiting reagent with regards to the Fe production? Fe2O3(s) + 3CO (g) —> 2Fe(s) + 3 CO2 (g)

Answer 1

Answer: CO is a limiting reagent with regards to the Fe production.

Step-by-step explanation:

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

Moles of CO =
`\frac{\text{mass of CO}}{\text{molar mass of CO}}=(13.2 g)/(28 g/mol)=0.4714 mol`

moles of
`Fe_2O_3=\frac{\text{mass of} Fe_2O_3}{\text{molar mass of}Fe_2O_3}=(42.7 g g)/(159.7 g/mol)=0.2673 mol`

According to reaction , 3 mole of CO reacts with 1 mole of
`Fe_2O_3` then , 0.4714 moles of CO will react with :
`(1)/(3)* 0.4714` moles of
`Fe_2O_3` that is 0.1571 moles.

0.4714 moles of CO will react with 0.1571 moles of
`Fe_2O_3` which means that CO is present in limited amount acting as limiting reagent.

Mole remaining of
`Fe_2O_3` = 0.2673 mol - 0.1571 mol = 0.1102 mol

Hence, CO is a limiting reagent and
`Fe_2O_3` is an excessive reagent.