Hydrazine, n2h4, may react with oxygen to form nitrogen gas and water. n2h4(aq) + o2(g) -> n2(g) + 2h2o(l)if 3.95 g of n2h4 reacts and produces 0.550 l of n2, at 295 k and 1.00 atm, what is the percent yield of the reaction?

The percent yield of the reaction is 18.45 %

Explanation:

The reaction is:

N₂H₄(aq) + O₂(g) →  N₂(g) + 2H₂O(l)

Ratio between hydrazine and N₂ is 1:1, so 1 mol of hydrazine produces 1 mol of N₂

The molar mass of hydrazine is 32  g/m

The moles we used are : mass / molar mass hydrazine

3.95 g / 32 g/m = 0.123 moles

So, in the 100 % yield reaction, we will produce 0.123 of gas.

Let's apply the Ideal Gases law to find out, the moles of gas that have been produced.

0.550L . 1 atm = n . 0.082 . 295K

(0.550L .1atm) / 0.082 . 295K = n

0.0227 mol = n

So, to find out the percent yield of the reaction we finally make a rule of three

0.123 moles ___ 100 %

0.0227 moles 18.45 %

the answer for sodium is 0.186

the answer for rubidium is 0.251