Hydranize, N2H4, may react with Oxygen to form nitrogen gas and water.

N2H4(aq) + O2(g) ----- N2(g) + 2H2O(l)

If 2.45 g of N2h4 reacts and produces 0.350 L of N2, at 295K and 1.00 atm, what is the percent yield of the reaction?

Question

Hydranize, N2H4, may react with Oxygen to form nitrogen gas and water. 

N2H4(aq) + O2(g) -----> N2(g) + 2H2O(l) 

If 2.45 g of N2h4 reacts and produces 0.350 L of N2, at 295K and 1.00 atm, what is the percent yield of the reaction?

photon336 · Answer

How would you approach this?

anonymous · Answer

i know i have to find the molar mass of n2h4 first... would that be .073?

anonymous · Answer

i know i have to find the molar mass of n2h4 first... would that be .073?

photon336 · Answer

$$yield = \frac{ experimental  }{ actual } * 100%$$

photon336 · Answer

Theoretical yield is always what you get from doing the stoichiometry.

we always work with our balanced equation first. we know our equation is balanced. so we can go right in

Well, the first thing we need to do is find out how many moles of N2H4 we have

$$2.45~grams~N_{2}H_{4}*(\frac{ mol }{ 32~grams })=\frac{ 2.45 }{ 32 } = 0.077~mol~N_{2}H_{2}$$

now, we take this number and multiply by the molar ratio to find the number of moles of N2 we would get from our formula right?

$$0.077~mol~N_{2}H_{4}*(\frac{ N_{2} }{ N_{2}H_{4} }) = 0.077~mol~N_{2}$$

since the molar ratio is 1/1 we expect to get 0.07 moles of N_{2} gas. 


$$0.077*24(\frac{ grams }{ mol }) =2.16~grams~N_{2}$$

photon336 · Answer

I like this problem. it's testing multiple concepts. 

now 2.16 grams of Nitrogen gas is how much we would expect if everything went as desired. this is our theoretical yield many times in lab you don't always get this number in fact you may get less than the desired number due to side reactions, plain loss of product, etc. 

our experimental yield is what you get by doing the experiment. in this case they've told you that you have 0.350 L of N_{2} but we can easily find the number of moles of this using the ideal gas equation. 

are you familiar with this equation? 

$$pV = nRT $$

photon336 · Answer

@deejayy

photon336 · Answer

$$\frac{ pV }{ RT } = n~moles$$

$$\frac{ (1atm)(0.350L) }{ (295K)(0.08Lmol^{-1}atmK^{-1}) } = n~moles~N_{2}$$

this is our experimental value. 

take the experimental value and place it over the theoretical value to get the yield.

anonymous · Answer

okay, i got it! thank you!!