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 %

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

yes

explanation:

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