Question

Ammonia reacts with oxygen according to the equation given below. If a `"72.0-mL"` sample of `"NH"_3` gas is allowed to react with excess oxygen at room temperature, `25^@"C"`, and `"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)`

Answer 1

`2.7 * 10^(21)`

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.