Question

How many grams of CO are needed to react with an excess of Fe2O3 to produce (210.3 g Fe) 209.7 g Fe? Show your work.

CO (g) + Fe2O3 (s) --> Fe (s) + CO2 (g)

Answer 1

5.625 moles. took the test

Answer 2

Step-by-step explanation:

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

1)Mass of CO when 210.3 g of Fe produced.

Number of moles of
`Fe` in 210.3 g=

`\frac{\text{Given mass of Fe}}{\text{Molar mass of Fe}}`

`=(210.3)/(55.84 g/mol)=3.76 mol`

According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.76 moles of Fe will be obtained from :
`(3)/(2)* 3.76 moles` of CO that is 5.64 moles.

Mass of CO in 5.64 moles =

`\text{Number of moles}* \text{molar mass of CO}=5.46* 28 g/mol=157.92 g`

2)Mass of CO when 209.7 g of Fe produced.

Number of moles of
`Fe` in 209.7 g=

`\frac{\text{Given mass of Fe}}{\text{Molar mass of Fe}}`

`=(209.7)/(55.84 g/mol)=3.75 mol`

According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.75 moles of Fe will be obtained from :
`(3)/(2)* 3.75 moles` of CO that is 5.625 moles.

Mass of CO in 5.625 moles =

`\text{Number of moles}* \text{molar mass of CO}=5.625* 28 g/mol=157.5 g`