Question #69942
1 Answer
Explanation:
The first order of business here is to figure out the number of moles of dinitrogen oxide produced by the reaction.
To calculate the number of moles of dinitrogen gas present in that
SIDE NOTE Keep in mind that most sources and textbooks still use a pressure of
In this case, one mole of any ideal gas occupies
In your case, the sample of dinitrogen oxide gas will contain
#0.240 color(red)(cancel(color(black)("L"))) * ("1 mole N"_2"O")/(22.7color(red)(cancel(color(black)("L")))) = "0.01057 moles N"_2"O"#
Now take a look at the balanced chemical equation that describes this decomposition reaction
#"NH"_ 4"NO"_ (3(s)) -> "N"_ 2"O"_ ((g)) + 2"H"_ 2"O"_ ((g))#
Notice that every mole of ammonium nitrate that undergoes decomposition produces
#0.01057 color(red)(cancel(color(black)("moles N"_2"O"))) * ("1 mole NH"_4"NO"_3)/(1color(red)(cancel(color(black)("mole N"_2"O")))) = "0.01057 moles NH"_4"NO"_3#
To convert this to grams, use the molar mass of ammonium nitrate
#0.01057 color(red)(cancel(color(black)("moles NH"_4"NO"_3))) * "80.043 g"/(1color(red)(cancel(color(black)("mole NH"_4"NO"_3)))) = color(darkgreen)(ul(color(black)("1.21 g")))#
The answer is rounded to three sig figs, the number of sig figs you have for the volume of dinitrogen oxide produced by the reaction.