Question

Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of NH3 reacts with 3.23 g O2 and produces 0.550 L of N2., at 295 K and 1.00 atm which reactant is limiting?NH3 or O2What is the percent yeild?

Answer 1

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

Step 1: Data given

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

Step 2: The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

Step 3: Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

Step 4: Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

Step 5: Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

Step 6: Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

Step 7: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

Step 8: Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

Answer 2

Ammonia (NH₃) is the limiting reactant

The percent yield is 36%

Stoichiometry

From the question,

We are to determine which reactant is limiting and the percent yield

First we will write the balanced chemical equation for the reaction

4NH₃(aq) + 3O₂(g) → 2N₂(g) + 6H₂O(l)

This means, 4 moles of ammonia reacts with 3 moles of oxygen gas to produce 2 moles of nitrogen gas and 6 moles of water

Now, we will determine the number of moles of each of the reactants present

• For ammonia, NH₃

Mass = 2.15 g

Using the formula,

$Number\ of\ moles = \frac{Mass}{Molar\ mass}$

Molar mass of ammonia = 17.031 g/mol

∴ Number of moles of ammonia present = $\frac{2.15}{17.031}$

Number of moles of ammonia present = 0.12624 mole

• For oxygen gas, O₂

Mass = 3.23 g

Molar mass of O₂ = 32.0 g/mol

∴ Number of moles of O₂ present = $\frac{3.23}{32.0}$

Number of moles of O₂ present = 0.1009375 mole

Since, 4 moles of ammonia is required to react with 3 moles of oxygen gas

Then,

0.12624 mole of ammonia will react with $\frac{0.12624\times 3}{4}$ moles of oxygen gas

$\frac{0.12624\times 3}{4} = 0.09468$

∴ The 0.12624 mole of ammonia will react with 0.09468 mole out of the  0.1009375 mole of oxygen present

This means oxygen is in excess.

Hence, Ammonia (NH₃) is the limiting reactant

Now, for the percent yield

Percent yield can be calculated using the formula,

$Percent \ yield =\frac{Actual\ yield }{Theoretical \ yield } \times 100\%$

Now, we will determine the theoretical yield

From the balanced chemical equation,

4 moles of ammonia reacts with 3 moles of oxygen gas to produce 2  moles of nitrogen gas

Then,

The 0.12624 mole of ammonia will react with 0.09468 mole os oxygen gas to produce 0.06312 mole of nitrogen gas

∴ The theoretical yield of nitrogen gas is 0.06312 mole

Now, we determine the actual yield in terms of number of moles

From the given information,

Actual yield in volume = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

From the ideal gas equation

PV= nRT

Then,

$n = \frac{PV}{RT}$

$n = \frac{1.00 \times 0.550}{0.08206 \times 295}$

$n = 0.02272 \ mole$

This is the actual yield in terms of number of moles

∴ $Percent\ yield = \frac{0.02272}{0.06312} \times 100\%$

Percent yield = 0.359949 × 100%

Percent yield = 35.9949%

Percent yield ≅ 36%

Hence, the percent yield is 36%

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