Ammonia and oxygen react to form nitrogen and water according to the following b

Question

Ammonia and oxygen react to form nitrogen and water according to the following balanced equation:

4NH3(g) + 3 O2(g) produces 2N2(g) + 6H2O(l)

a. Calculate the mass of "one"mole of each of the reactants and products in the above reaction.

   NH3 O2 N2 H2O

b. Calculate the moles of oxygen required to produce 8.50 moles of water.

c. How many grams of nitrogen will be produced when 4.35 moles of ammonia react completely?

d. How many grams of water will be produced when 26.00 grams of oxygen react completely?

***PLEASE show how you arrived at your answers for future problem calculations, Thank you! Need these answered ASAP***

Explanation / Answer

a) consider the given reaction

4NH3 + 302 -----> 2N2 + 6H20

one moles of NH3 = 17 g

one mole of 02 = 32 g

one moles of N2 = 28 g

one mole of H20 = 18 g

b)

now

we can see that

moles of H20 produced = 2 x moles of O2

soo

8.5 = 2 x moles of 02

moles of 02 = 4.25

so

4.25 moles of oxygen is required

c)

we can see that

moles of N2 produced = 0.5 x moles of NH3 reacted

so

moles of N2 produced = 0.5 x 4.35

moles of N2 produced = 2.175

now

mass = moles x molar mass

so

mass of N2 = 2.175 x 28

mass of N2 = 60.9

so

60.9 grams of nitrogen will be produced

d)

moles = mass / molar mass

moles of 02 = 26 /32

now

moles of H20 produced = 2 x moles of 02

moles of H20 produced = 2 x 26 /32 = 52 /32

now

mass of H20 produced = 52 x 18 /32

mass of H20 produced = 29.25

so

29.25 grams of H20 is produced

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