How many molecules of water will be generated by the complete reaction of 191.4 grams of ammonia?
balanced equation: """4 nh3 (g) + 7 o2 (g) => 4 no2 (g) + 6 h2o (g)"""
answer key: 1.015 x 10 ^ 25 molecules h2o
> i am not getting that 2.5416 x 10 ^ 24
> what's the correct work needed to be done to get the right answer?

To find:

Number of water molecules needed to generate 191.4 grams of ammonia.

So, firstly write the molecular mass of ammonia = 14 + 3 = 17 grams

1 mol of ammonia= 17 grams

1 gram of ammonia= moles of ammonia

191.4 grams of ammonia =

From the balanced equation,

4 NH₃ (g) + 7O₂ (g) → 4 NO₂ (g) + 6 H₂O (g)

It is know that:

4 moles of ammonia produces 6 moles of water.

Then ,11.25 moles of ammonia will produce

From Avogadro's number:

1 mole of water =  6.023 x 10 ²³ molecules of water

11.25 moles of ammonia is producing 16.875 moles of water.

Therefore,

Number of water molecules in 11.25 moles of ammonia will be:

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The answer to your problem is: 1.015 x 10²⁵ molecules of water

Explanation:

We know that were generated 191.4 grams of ammonia, but we need to convert this value to moles, in this way it will be easier to solve the problem.

Then molecular mass of ammonia = 14 + 3 = 17 grams

    17 grams of ammonia 1 mol

  191.4 grams    x

x = 191.4 x 1 / 17 = 11.25 moles

From the balanced equation we know that

                 4 moles of ammonia  6 moles of water

Then        11.25 moles of ammonia      x

x = 11.25 x 6 / 4 = 16.875 moles of water

Finally using Avogadro's number

            6.023 x 10 ²³ molecules of water 1 mole of water

                      x           molecules of water   11.25 moles of water

x = 11.25 x ( 6.023 x 10 ²³) / 1 = 1.015 x 10 ²⁵ molecules of water

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