# 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.

Aug 3, 2017

The actual yield using the molar volume $\text{22.414 mol/L}$ is $\text{7.88 L N"_2}$.

The actual yield using the molar volume $\text{22.71 mol/L}$ is $\text{7.97 L N"_2}$

#### Explanation:

Balanced Equation

$\text{NH"_4"NO"_2}$$\rightarrow$$\text{N"_2" + 2H"_2"O}$

Determine moles $\text{NH"_4"NO"_2}$. The molar mass of ammonium nitrite is $\text{64.044 g/mol}$.

We will need the molar volume of a gas for these calculations. Molar volume at the most commonly used values for STP, ${0}^{\circ} \text{C}$, or $\text{273.15 K}$ (for gases), and $\text{1 atm}$. The molar volume of a gas under these conditions is $\text{22.414 L/mol}$.

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 $\text{N"_2}$ produced by multiplying the moles ${\text{NH"_4"NO}}_{2}$ by the mole ratio between ${\text{N}}_{2}$ and $\text{NH"_4"NO"_2}$ from the balanced equation. $\text{1 mol N"_2} :$ $\text{1 mol NH"_4"NO"_2}$

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 $\text{N"_2}$ produced by multiplying by the mole $\text{N"_2}$ by the molar volume of a gas $\left(\text{22.414 mol/L}\right)$.

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 ${\text{N}}_{2}$ by the percent yield.

$17.5 \text{ L N"_2xx0.450="7.88 L N"_2}$

The actual yield is $\text{7.88 L N"_2}$.

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The most current values for STP, recommended by the International Pure and Applied Chemistry (IUPAC) in 1982, are ${0}^{\circ} \text{C}$, or $\text{273.15 K}$ and ${10}^{5} \text{ Pa}$. Under these conditions, The molar volume of a gas is $\text{22.71 L/mol}$.

Theoretical yield

$0.7807 \text{mol N"_2xx(22.71"L N"_2)/(1"mol N"_2)="17.7 L N"_2}$

Actual yield

$17.7 \text{L N"_2xx0.450="7.97 L N"_2}$