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

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

If 2.05g of N2H4 reacts and produces 0.550 L of N2, at 295K and 1.00 atm, what is the percent yield of the reaction?

Respuesta :

anfabba15

Answer:

The percent yield of the reaction is 35 %

Explanation:

In the reaction, 1 mol of hydrazine reacts with 1 mol O₂ to produce 1 mol of nitrogen and 2 moles of water.

Let's verify the moles that were used in the reaction.

2.05 g . 1mol/ 32 g = 0.0640 mol

In the 100% yield, 1 mol of hydrazine produce 1 mol of N₂ so If I used 0.0640 moles of reactant, I made 0.0640 moles of products.

Let's use the Ideal Gases Law equation to find out the real moles of nitrogen, I made (real yield).

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

(1atm . 0.550L) / 0.082 . 295K = n → 0.0225 moles

Percent yield of reaction = (Real yield / Theoretical yield) . 100

(0.0225 / 0.0640) . 100 = 35%

Eduard22sly

Answer: 35.5%

Explanation:

First we need to find the number of mole of 2.05g of N2H4

Molar Mass of N2H4 = (14x2) + (1x4)

= 28 + 4 = 32g/mol

Mass conc. of N2H4 = 2.05g

Number of mole = Mass conc /Molar Mass

Number of mole = 2.05/32 = 0.0641mol

Next, We need to find the volume (theoretical yield) occupied by this mole(0.0641mol) of N2H4, using the ideal gas equation.

n = 0.0641mol

P = 1atm

T = 295K

R = 0.082atm.L/mol /K

V =?

PV = nRT

1 x V = 0.0641 x 0.082 x 295

V = 1.5506L

The volume (1.5506L) obtained is the theoretical yield

But the experimental volume = 0.550 L

Percentage yield =( Experimental yield / theoretical yield) x100

= (0.550/1.5506) x 100

= 35.5%