Ammonia gas is formed from nitrogen gas and hydrogen gas, according to the following equation, n2 (g) + 3h2 (g) 2nh3 (g). if 140 grams of nitrogen gas is allowed to react with an excess of hydrogen gas to produce 155 grams of ammonia, what is the percent yield of this reaction?

The percent yield of the reaction is 91.20 %.

Explanation:

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

....(1)

Given mass of nitrogen gas = 140 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in above equation, we get:

For the given chemical reaction:

As, hydrogen is present in excess. Thus, it is considered as an excess reagent. Nitrogen gas is considered as limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of nitrogen gas produces 2 moles of ammonia.

So, 5 moles of nitrogen gas will produce = of ammonia.

Now, calculating the mass of ammonia from equation 1, we get:

Molar mass of ammonia = 17 g/mol

Moles of ammonia = 10 moles

Putting values in above equation, we get:

To calculate the percentage yield of the reaction, we use the equation:

Experimental yield of ammonia = 155 g

Theoretical yield of ammonia = 170 g

Putting values in above equation, we get:

Hence, the percent yield of the reaction is 91.20 %.

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

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