chapter 7 gen chem

Question

Hydrazine, N2H4, reacts with oxygen to form nitrogen gas and water.

N2H4(aq) + O2(g) ------ N2(g)+2H2O(1)

If  2.15g of N2H4 reacts with excess oxygen and produces 0.750L of N2, at 295K and 1.00atm what is the percent yield of the reaction?

percent yield: =%

Answer 1

Hello, Sarah,

This is a 2 -part problem. Thanks for providing a balanced equation - it saves time. We need to determine the amount of N₂ produced, but all we are given are the volume of N₂ under specified conditions of temperature and pressure. To find how many of moles (and then grams) of N₂ we need the ideal gas law, PV = nRT.

Rearrange for n, the moles of gas: n = PV/RT.

I chose a value of the gas constant, R, that is most compatible with the units of the data provided: 0.0820575 L⋅atm⋅K⁻¹⋅mol⁻¹. Enter the data, cancel units, and I found n = 0.0310 moles.

0.0310 moles of N₂, when converted to grams by multiplying by the molar mass of N₂ (28 g/mole) tells us we produced all of 1.36 grams of nitrogen gas.

Now we need to use the balanced equation to determine what we would expect to produce, if everything went according to plan (theory). The equation states that we should expect to produce 1 mole of N₂ for every 1 mole of N₂H₄. 2.15 grams of hydrazine, with a molar mass of 32 grams/mole, tells us we have 0.0672 moles of N₂H₄. Therefore, we'd expect the same number of moles of N₂. 0.0672 moles of N₂ times it's molar mass of 28 grams/mole, means we should have 1.88 grams of N₂.

We actually obtained 1.36 grams, so the percent theorectical yield is 1.36/1.88 or 72.3%. Not good, not bad, but a lot of work for only 1.36 grams of a very common gas. Next time, pick a gas that's more fun, such as HeHeHeHe.

Bob