Ammonia reacts with oxygen according to the equation given below. If a "72.0-mL"72.0-mL sample of "NH"_3NH3 gas is allowed to react with excess oxygen at room temperature, 25^@"C"25∘C, and "1 atm"1 atm, calculate the number of molecules of water produced?

4"NH"_3(g) + 5"O"_2(g) -> 4"NO"(g) + 6"H"_2"O"(l)4NH3(g)+5O2(g)→4NO(g)+6H2O(l)

1 Answer
Stefan V.
Jul 9, 2017

2.7 * 10^(21)2.7⋅1021

Explanation:

The first thing that you need to do here is to use the ideal gas law equation

color(blue)(ul(color(black)(PV = nRT)))

Here

  • P is the pressure of the gas
  • V is the volume it occupies
  • n is the number of moles of gas present in the sample
  • R is the universal gas constant, equal to 0.0821("atm L")/("mol K")
  • T is the absolute temperature of the gas

to find the number of moles of ammonia present in your sample.

Rearrange the ideal gas law equation to solve for n

PV = nRT implies n = (PV)/(RT)

Plug in your values to find--do not forget that you need to convert the volume of the sample to liters and the temperature to Kelvin!

n = (1 color(red)(cancel(color(black)("atm"))) * 72.0 * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K"))))

n= "0.0029414 moles NH"_3

Now, you know by looking at the balanced chemical equation

4"NH"_ (3(g)) + 5"O"_ (2(g)) -> 4"NO"_ ((g)) + 6"H"_ 2"O"_ ((l))

that for every 4 moles of ammonia that take part in the reaction you get 6 moles of water.

This means that your reaction will produce--oxygen gas is said to be in excess, so you don't have to worry about it being a limiting reagent.

0.0029414 color(red)(cancel(color(black)("moles NH"_3))) * overbrace(("6 moles H"_2"O")/(4color(red)(cancel(color(black)("moles NH"_3)))))^(color(blue)("given by the balanced chemical equation"))

= "0.004412 moles H"_2"O"

To find the number of molecules of water produced by the reaction, use the fact that it takes 6.022 * 10^(23) molecules of water in order to have exactly 1 mole of water -> this is given by Avogadro's constant.

You can thus say that your sample will contain

0.004412 color(red)(cancel(color(black)("moles H"_2"O"))) * overbrace((6.022 * 10^(23)color(white)(.)"molecules H"_2"O")/(1color(red)(cancel(color(black)("mole H"_2"O")))))^(color(blue)("Avogadro's number"))

= color(darkgreen)(ul(color(black)(2.7 * 10^(21)color(white)(.)"molecules H"_2"O")))

I'll leave the answer rounded to two sig figs, but don't forget that you only have one significant figure for the pressure at which the reaction takes place.

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