## Question M.1 [ENDORSED]

Posts: 62
Joined: Thu Jul 27, 2017 3:01 am

### Question M.1

Hydrazine, N2H4, is an oily liquid used as a rocket fuel. It
can be prepared in water by oxidizing ammonia with hypochlorite
ions: 2 NH31g2  ClO1aq2 S N2H41aq2  Cl1aq2  H2O1l2.
When 35.0 g of ammonia reacted with an excess of hypochlorite
ion, 25.2 g of hydrazine was produced. What is the percentage
yield of hydrazine?

I understand that I have to find the grams of hydrazine eventually. Is that the theoretical or actual yield?

K Stefanescu 2I
Posts: 68
Joined: Fri Sep 29, 2017 7:04 am

### Re: Question M.1  [ENDORSED]

You are correct that you must calculate the mass of hydrazine produced from 35.0g of ammonia. You will find that this resultant theoretical yield does not match the 25.2g of hydrazine actually produced. Thus, you must determine percent yield by dividing the actual yield (given) by the theoretical yield (calculated), and then multiplying by 100 to obtain a percentage.

Michelle Dong 1F
Posts: 110
Joined: Fri Sep 29, 2017 7:04 am

### Re: Question M.1

Generally, the theoretical yield is the one you're supposed to calculate, and the actual yield is the one already given to you by the problem.

Elizabeth Bamishaye 2I
Posts: 54
Joined: Fri Sep 29, 2017 7:04 am

### Re: Question M.1

Basically, you are solving for the theoretical yield and once you get that you divide it from the 25.2g of hydrazine produced then multiply by 100 to receive the percentage yield of hydrazine.

Nehal Banik
Posts: 64
Joined: Thu Jul 13, 2017 3:00 am

### Re: Question M.1

Since you already know that hypochlorite is in excess, you assume that ammonia is the limiting reactant, therefore convert the 35g of ammonia to grams of hydrazine and then you will find the theoretical yield. Once you find the theoretical yield, you take the actual yield which is 25.2 grams produced and divide by the theoretical grams of hydrazine and you multiply by 100, your answer will be the percentage yield of hydrazine.

Nyari Muchaka_Discussion 4A
Posts: 53
Joined: Thu Sep 19, 2019 12:17 am

### Re: Question M.1

Since ammonia is the limiting reactant you have to convert from moles of ammonia to moles of hydrazine.
1. 35g/17.04(molar mass of ammonia)=2.054mol of ammonia.
Since ammonia and hydrazine are in a two to one ratio you use the mol of ammonia to calculate the mol fo hydrazine.
2. 2.054/2 is 1.0269 mol hydrazine. This represents the theoretical yield of hydrazine as it is how much should be produced based on the balanced equation.
Now find the mol of hydrazine using the molar mass and the given grams of hydrazine.
3. 25.2/32.0452-.78015 mol hydrazine. This represent the actual yield of hydrazine as it was the amount actually produced.
4. (.78015/1.0269)x100=76.6%