Balanced Equation
#"NH"_4"NO"_2"##rarr##"N"_2" + 2H"_2"O"#
Determine moles #"NH"_4"NO"_2"#. The molar mass of ammonium nitrite is #"64.044 g/mol"#.
We will need the molar volume of a gas for these calculations. Molar volume at the most commonly used values for STP, #0^@"C"#, or #"273.15 K"# (for gases), and #"1 atm"#. The molar volume of a gas under these conditions is #"22.414 L/mol"#.
#50.0color(red)cancel(color(black)("g NH"_4"NO"_2))xx(1"mol NH"_4"NO"_2)/(64.044color(red)cancel(color(black)("g NH"_4"NO"_2)))="0.7807 mol NH"_4"NO"_2#
I am keeping an extra digit to reduce rounding errors. I will round to three significant figures at the end.
Determine moles #"N"_2"# produced by multiplying the moles #"NH"_4"NO"_2# by the mole ratio between #"N"_2# and #"NH"_4"NO"_2"# from the balanced equation. #"1 mol N"_2":# #"1 mol NH"_4"NO"_2"#
#0.7807color(red)cancel(color(black)("mol NH"_4"NO"_2))xx(1"mol N"_2)/(1color(red)cancel(color(black)("mol NH"_4"NO"_2)))="0.7807 mol N"_2"#
Determine the theoretical volume of #"N"_2"# produced by multiplying by the mole #"N"_2"# by the molar volume of a gas #("22.414 mol/L")#.
#0.7807color(red)cancel(color(black)("mol N"_2))xx(22.414"L N"_2)/(1color(red)cancel(color(black)("mol N"_2)))="17.5 L N"_2"#
Convert percent yield to decimal form.
#"Percent yield"=45.0%=45.0/100=0.450#
Determine actual yield.
Multiply the theoretical volume of #"N"_2# by the percent yield.
#17.5" L N"_2xx0.450="7.88 L N"_2"#
The actual yield is #"7.88 L N"_2"#.
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The most current values for STP, recommended by the International Pure and Applied Chemistry (IUPAC) in 1982, are #0^@"C"#, or #"273.15 K"# and #10^5" Pa"#. Under these conditions, The molar volume of a gas is #"22.71 L/mol"#.
Theoretical yield
#0.7807"mol N"_2xx(22.71"L N"_2)/(1"mol N"_2)="17.7 L N"_2"#
Actual yield
#17.7"L N"_2xx0.450="7.97 L N"_2"#
