Fundamentals 7th Edition M1

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rikolivares
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Joined: Fri Sep 28, 2018 12:23 am

Fundamentals 7th Edition M1

Postby rikolivares » Thu Oct 04, 2018 11:55 am

The problem asks

Hydrazine, N 2 H 4, is an oily liquid used as a rocket fuel. It can be prepared in water by oxidizing ammonia with hypochlorite ions: 2 NH3(g) + ClO (aq) = N2H4(aq) + Cl (aq) + H2O(l). 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 followed the instructions on page F97 and I'm pretty sure I got all my calculations right but I am still getting a percentage that is over 100%, which obviously is not possible. Am I using the wrong method by using the limiting reagent to ratio hydrazine?

Julia Go 2L
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Joined: Sun Sep 30, 2018 12:17 am
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Re: Fundamentals 7th Edition M1

Postby Julia Go 2L » Thu Oct 04, 2018 12:43 pm

Hello! To calculate the percent yield, you use the equation:

percent yield = (actual yield/theoretical yield) * 100.

In this problem, the question gives you the information that the actual yield is 25.2g of hydrazine.

All that's left to calculate is the theoretical yield of N2H4 and plug it into the equation.

To find this, first note that the question tells you that hypochlorite is in "excess," which means that ammonia is the limiting reactant. Calculate the amount of moles of the limiting reactant by the dividing the 35.g of ammonia by the molar mass of ammonia.

35.0g / 17.031g = 2.055 moles NH3

Next, look back at the balanced equation and notice that it takes 2 moles of NH3 to produce 1 mole of N2H4. (ratio of 2:1) So:

2.055 moles NH3 * (1 mole N2H4/ 2 moles NH3) = 1.0275 mol N2H4

Finally, convert the result back to grams. Multiply 1.0275 mol by the molar mass of N2H4. Then you should get the theoretical yield of N2H4 and you can plug that value into the percent yield equation.

Anita Wong 1H
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Re: Fundamentals 7th Edition M1

Postby Anita Wong 1H » Thu Oct 04, 2018 12:45 pm

As per what the instructions on page F97 say:


Since we are given the actual yield (25.2 g N2H4), we need to calculate the theoretical yield.


1) Start with the given mass of the reactant: 35.0 g NH3

2) Divide the mass by the molar mass of reactant NH3 to get the amount (mol) of reactant: 35.0 g NH3 / 17.0337 g.mol NH3 = 2.054750289 mol NH3

3) To go from 2NH3 (reactant) to N2H4 (product) , we use the mole ratio for the reaction (2 NH3 : 1 N2H4).
- Divide 2 from 2.054750289 mol NH3 (we do this because we want to go from 2 moles (from NH3) to 1 mol ( for N2H4):
2.054750289 mol NH3 / 2 mol NH3 = 1.027375145 mol N2H4

4) Now that we have the moles of N2H4, we use the molar mass of the product (N2H4) to get the theoretical yield in grams:
- Multiply 1.027375145 mol N2H4 by the molar mass of N2H4: 1.027375145 mol N2H4 x 32.0516 g.mol = 32.9 g N2H4 <----- Theoretical yield


Time to calculate the percentage yield.

% yield = actual yield / theoretical yield x 100%

% yield of N2H4 = 25.2 / 32.9 x100% = 76.6%


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