Question

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 answer: 2.5416 x 10 ^ 24
> What's the correct work needed to be done to get the right answer?

Answer 1

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

Step-by-step 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