9. 6.59 g of ammonia (nh3, 17.031 g/mol) react completely in the presence of excess carbon dioxide (co2, 44.01 g/mol) according to the following balanced equation,
2nh3(g) + co2(g) → ch4n2o(s) + h2o(l) if 7.40 g of urea (ch4n2o, 60.056 g/mol) are produced,
calculate the percent yield for the reaction.

The percent yield for the reaction is 63.7%

Explanation:

This problem can be solved, from different rules of three.

The reaction is:

2NH₃(g) + CO₂(g) → CH₄N₂O(s) + H₂O(l)

2 moles of ammonia, produce 1 mol of urea.

Mass / molar mass = moles

6.59g / 17.031 g/m = 0.387 moles

As ratio is 2:1, I will produce half a mole of urea.

(0.387 m . 1m)/ 2m = 0.193 m

Let's find out the mass.

Mass of urea = Moles of urea . molar mass of urea.

Mass = 0.193m .  60.056 g/m = 11.6 g

If the percent yield for the reaction was 100 %, we make 11.6 g, but I only obtained 7.40 g so let's calculate the new percent yield.

11.6 g 100 %

7.40 g ___ (7.40g / 11.6g) .100 = 63.7 %

it is important to conserve water because there is very limited amount of freshwater here on earth and if we are not smart about how we use it we will only be runnning out of our supply faster than neccessarry.

answer: your mm

explanation: