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 %

indium, and thallium have three electrons i think

let that person have a turn to speak : )