Question

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

N2H4(aq)+O2(g)⟶N2(g)+2H2O(l) If 3.15 g of N2H4 reacts with excess oxygen and produces 0.950 L of N2 , at 295 K and 1.00 atm, what is the percent yield of the reaction?

Answer 1

Percent yield = 39.3%

Step-by-step explanation:

Given data:

Mass of N₂H₄ = 3.15 g

Volume of nitrogen produced = 0.950 L

Temperature = 295 K

Pressure = 1 atm

Percent yield = ?

Solution:

Chemical equation:

N₂H₄ + O₂ → N₂ + 2H₂O

Number of moles of N₂H₄:

Number of moles = mass/ molar mass

Number of moles = 3.15 g/32 g/mol

Number of moles = 0.1 mol

Now we will compare the moles of hydrazine and nitrogen.

N₂H₄ ; N₂

1 : 1

0.1 : 0.1

PV = nRT

V = nRT/P

V = 0.1 mol ×0.0821 atm.L/mol.K ×295 K / 1 atm

V = 2.42 L

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 0.950 L/ 2.42 L × 100

Percent yield = 39.3%