Ammonia (NH3) can be produced by the reaction of hydrogen gas with nitrogen gas:

3H2 + N2→2NH3

A chemist reacts 2.00 mol H2 with excess N2. The reaction yields 0.13 mol NH3.

What is the percent yield of the reaction?

10%

40%

60%

80%

10 percent is correct

10%

Explanation:

Given dа ta:

Actual yield = 0.13 moles

Percent yield = ?

Solution:

Balanced chemical equation.

3H₂ + N₂ → 2NH₃

First of all we will calculate the mass of given moles of ammonia.

Number of moles = mass / molar mass

Mass = 0.13 * 17 g/mol

Mass = 2.21 g

2.21 g is experimental yield of ammonia.

Hydrogen is limiting reactant because only two moles of hydrogen present.

we will compare the moles of hydrogen and ammonia from balance chemical equation.

H₂ : NH₃

3 : 2

2 : 2/3*2 = 1.33 moles

Mass of ammonia

Mass = number of moles * molar mass

Mass = 1.33 mol * 17 g/mol

Mass = 22.61 g

Percent yield:

Percent yield = actual yield / theoretical yield * 100

Percent yield = 2.21 g / 22.61 g

Percent yield = 9.8% which is almost 10%