What volume of nitrogen gas can be produced by the decomposition of 50.0 g of ammonium nitrite if the yield for this reaction is 45.0%? Water is the other product.

1 Answer
Aug 3, 2017

Answer:

The actual yield using the molar volume #"22.414 mol/L"# is #"7.88 L N"_2"#.

The actual yield using the molar volume #"22.71 mol/L"# is #"7.97 L N"_2"#

Explanation:

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"#