Hydrazine, n2h4 , reacts with oxygen to form nitrogen gas and water. n2h4(aq)+o2(g)⟶n2(g)+2h2o(l) if 2.65 g of n2h4 reacts with excess oxygen and produces 0.350 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 17.41 %.

Explanation:

To calculate the number of moles, we use the equation:

Given mass of = 2.65 g

Molar mass of = 32.04 g/mol

Putting values in above equation, we get:

To calculate the number of moles, we use the equation given by ideal gas equation:

where,

P = pressure of the gas = 1 atm

V = Volume of gas = 0.350 L

n = Number of moles = ?

R = Gas constant =

T = temperature of the gas = 273 K

Putting values in above equation, we get:

Now, to calculate the experimental yield of , we use the equation:

  .....(1)

We are given:

Moles of nitrogen gas = 0.0144 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in above equation, we get:

Experimental yield of nitrogen gas = 0.4032 g

By Stoichiometry of the reaction:

1 mole of produces 1 mole of nitrogen gas.

So, 0.0827 moles of will produce = of nitrogen gas.

Now, to calculate the theoretical yield of nitrogen gas, we use equation 1:

Moles of nitrogen gas = 0.0827 mol

Molar mass  nitrogen gas = 28 g/mol

Putting values in above equation, we get:

Theoretical yield of nitrogen gas = 2.3156 g

Experimental yield of nitrogen gas = 0.4032 g

Theoretical yield of nitrogen gas = 2.3156 g

Putting values in above equation, we get:

Hence, the percent yield of the reaction is 17.41 %.

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.