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 data:

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%

10%

Explanation:

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