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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the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.