A 3.00-L flask is filled with gaseous ammonia, NH3. The gas pressure measured at 20.0 ∘C is 2.55 atm . Assuming ideal gas behavior, how many grams of ammonia are in the flask?

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Answer 1 · 2018-01-24T06:27:42

$5.45$ $5.45$ grams of ammonia

Well, we would need to find the number of moles of ammonia $(N H_{3})$ $(NH_3)$ from the ideal gas law, and then we can find the mass.

The ideal gas formula states that

$P V = n R T$ $PV=nRT$

$P$ $P$ is pressure in atms (for this case)

$V$ $V$ is the volume in liters (for this case)

$n$ $n$ is the moles of the substance

$R$ $R$ is the ideal gas constant (varies, depending on other factors)

$T$ $T$ is temperature in Kelvins.

Now, we need to do some conversions.

$20^{\circ} C = 293.15 K$ $20^@C = 293.15K$

$n = \frac{P V}{R T}$ $n=(PV)/(RT)$ (since we need to find moles)

Since pressure is in atms, volume in liters, we use $R=0.082057 L \ atm \ mol^-1 \ K^-1$

Now, we can setup the equation.

$n=(2.55atm*3.00L)/(0.082057 \ L \ atm \ mol^-1 \ K^-1*293.15K)$

Also here, it is important to cancel units as well.

$n=(2.55cancel(atm)*3.00cancelL)/(0.082057 \ cancelL cancel(atm) \ mol^-1 \ cancel(K^-1)*293.15cancelK)$

$n=7.65/(24.06mol^-1)$

$n~~0.32mol$

So, there exists $0.32$ moles of ammonia $(NH_3)$ . To find the mass, we need to find the molar mass of ammonia and multiply it by the moles.

$"mass"="molar mass"*"moles"$

Ammonia has a molar mass of $17.031g"/"mol$ .

$:."mass"= 17.031g"/"cancel(mol)*0.32cancel(mol)$

$"mass"=5.45g$