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?

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.