Question

What is the theoretical yield of ammonia (in grams) if 16.55 grams of nitrogen gas and 10.15 grams of hydrogen gas are allowed to react?

Answer 1

Answer : 20.1 grams

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

For nitrogen:

$\text{Moles of nitrogen gas}=\frac{16.55g}{28g/mol}=0.591mol$

For hydrogen gas:

$\text{Moles of hydrogen gas}=\frac{10.15g}{2g/mol}=5.075mol$

The chemical equation for the reaction is:

$N_2+3H_2\rightarrow 2NH_3$

By Stoichiometry of the reaction:

1 mole of nitrogen gas reacts with 3 moles of hydrogen.

So, 0.591 moles of nitrogen gas will react with = $\frac{3}{1}\times 0.591=1.77mol$ of hydrogen.

As, given amount of hydrogen is more than the required amount. So, it is considered as an excess reagent.

Thus, nitrogen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of nitrogen gas produces 2 moles of ammonia

So, 0.591 moles of  nitrogen gas will produce = $\frac{2}{1}\times 0.591=1.18moles$ of ammonia

Now, calculating the mass of ammonia t:

$1.18mol=\frac{\text{Mass of ammonia}}{17g/mol}\\\\\text{Mass of ammonia}=20.1g$

Thus  theoretical yield of ammonia is 20.1 grams

Answer 2

Answer:

The theoretical yield of ammonia is 20.13 grams

Explanation:

Step 1: Data given

Mass of N2= 16.55 grams

Mass of H2 = 10.15 grams

Molar mass of N2 = 28 g/mol

Molar mass of H2 = 2.02 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate moles N2

Moles N2 = mass N2 / molar mass N2

Moles N2 = 16.55 grams / 28.0 g/mol

MolesN2 = 0.591 moles

Step 4: Calculate moles H2

Moles H2 = 10.15 grams / 2.02 g/mol

Moles H2 = 5.02 moles

Step 5: Calculate limiting reactant

For 1 mol N2 we need 3 moles H2 to produce 2 moles of NH3

N2 is the limiting reactant. It will completely be consumed (0.591 moles).

H2 is in excess. There will reat 3*0.591 = 1.773 moles

There will remain 5.02 - 1.773 = 3.247 moles H2

Step 6: Calculate moles of NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles of NH3

For 0.591 moles N2 we'll have 2-0.591 = 1.182 moles NH3

Step 7: Calculate mass of NH3

Mass NH3 = moles NH3 * molar mass NH3

Mass NH3 = 1.182 * 17.03 g/mol

Mass NH3 = 20.13 grams

The theoretical yield of ammonia is 20.13 grams