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.

Answer 1

The balanced chemical reaction is given as:

3CO (g) + Fe2O3 (s) --> 2Fe (s) + 3CO2 (g)

We are given the amount of the product to be produced from the reaction. We use this as the starting value for the calculations. We do as follows:

209.7 g Fe ( 1 mol / 210.3 g Fe ) ( 3 mol CO / 2 mol Fe ) (28.01 g / 1 mol) = 41.90 g CO needed

Answer 2

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=

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

$=\frac{210.3}{55.84 g/mol}=3.76 mol$

According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.76 moles of Fe will be obtained from : $\frac{3}{2}\times 3.76 moles$ of CO that is 5.64 moles.

Mass of CO in 5.64 moles =

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

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

Number of moles of $Fe$ in 209.7 g=

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

$=\frac{209.7}{55.84 g/mol}=3.75 mol$

According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.75 moles of Fe will be obtained from : $\frac{3}{2}\times 3.75 moles$ of CO that is 5.625 moles.

Mass of CO in 5.625 moles =

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