Question

For the reaction 3 CO + Fe2O3 --> 2 Fe + 3 CO2, how many liters of carbon monoxide at STP are needed to produce 1,001 grams of metallic iron?

Answer 1

Answer:

600.6L

Explanation:

Step 1:

The balanced equation for the reaction.

3CO + Fe2O3 --> 2Fe + 3CO2

Step 2:

Determination of the number of moles in 1001g of Fe.

This is illustrated below:

Mass of Fe = 1001g

Molar Mass of Fe = 56g/mol

Number of mole =?

Number of mole = Mass/Molar Mass

Number of mole of Fe = 1001/56

Number of mole of Fe = 17.875 mole

Step 3:

Determination of the number of mole of CO that reacted during the process.

This is illustrated below:

3CO + Fe2O3 --> 2Fe + 3CO2

From the balanced equation above,

3 moles of CO produced 2 moles of Fe.

Therefore, Xmol of CO will produce 17.875 moles of Fe i.e

Xmol of CO = (3 x 17.875)/2

Xmol of CO = 26.8125 moles

Step 4

Determination of the volume occupied by 26.8125 moles of CO at stp.

1 mole of a gas occupy 22.4L at stp.

Therefore 26.8125 moles of CO will occupy = 26.8125 x 22.4 = 600.6L

Therefore, 600.6L of CO is needed to produce 1001g of Fe

Answer 2

Answer:

$V=602L$

Explanation:

Hello,

In this case, for the given chemical reaction:

$3 CO + Fe_2O_3 \rightarrow 2 Fe + 3 CO_2$

In such a way, for the 1,001 g of iron, we compute the required moles of carbon monoxide by using the 2:3 mole ratio:

$n_{CO}=1,001gFe*\frac{1molFe}{55.845gFe}*\frac{3molCO}{2molFe} \\\\n_{CO}=26.89molCO$

Finally, we use the ideal gas equation to compute the volute at STP conditions (1atm and 273K):

$V=\frac{nRT}{P}=\frac{26.89mol*0.082\frac{atm*L}{mol*K}*273K}{1atm} \\\\V=602L$

Best regards.