What volume (in L) of nitrogen monoxide will be required to produce 54.5 L of carbon dioxide in

the reaction below?

2CO(g) + 2NO(g) → N2(g) + 2CO2(g)

54.5 L of nitrogen monoxide, NO.

Explanation:

The balanced equation for the reaction is given below:

2CO + 2NO —> N₂ + 2CO₂

From the balanced equation above,

2 L of NO reacted to produce 2 L of CO₂.

Finally, we shall determine the volume of nitrogen monoxide, NO, required to produce 54.5 L of carbon dioxide, CO₂. This can be obtained as follow:

From the balanced equation above,

2 L of NO reacted to produce 2 L of CO₂.

Therefore, 54.5 L of NO will also react to produce 54.5 L of CO₂.

Thus, 54.5 L of nitrogen monoxide, NO, is needed for the reaction.

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

the answer is b. dilution