Calculate the mass of 15.0 L of NH3 at 27 degrees C and 900 mmHg.

Mass of NH3 = 12.2 g

Mine doesn't round up. What did I do wrong?

15.0 L
27 degrees C + 273 = 300K
900. mmHg / 760 mmHg = 1.18 atm

1.18 atm x 15.0 L = n (.08206 Latm / mol K) 300K
n = .719 moles NH3 x (17.03056g / 1 moleNH3) = 12.2449 g NH3

The difference:

If I use 900/760 and round that number to 1.18 (which is correct), I get 12.22 which rounds to 12.2.
However, if I set my calculator to read 6 places (which I do routinely for all of my work) then go through without taking anything out of the calculator; ie., all of it done in one step, I get 12.266 which rounds to 12.3. Doing it your way causing rounding errors and this is an example. It becomes worse in problems where you take the number out of the calculator several times. Sometimes those rounding errors cancel, sometimes they don't. I think they don't most of the time. Besides, I don't like to take the time to read the calculator and copy the number onto a sheet of paper, then I must re-enter it into the calculator. All of that takes extra time and I'm too lazy to do that. By the way, many read that number (before taking it outof the calculator) to ONE MORE PLACE than allowed (you would have read 1.184) then when you re-entered it you would get the answer I get (almost) of 12.26 THEN you round to the correct number of s.f. which would be 12.3. Hope this digression helps make sense.

My calculator reads 12.267 so I would round that to 12.3 g.

It does. Thank you so much for all your help.

good. Come back anytime.

Why did the ammonia go on a diet? Because it wanted to be a lighter gas! But in all seriousness, to calculate the mass of NH3, we can use the ideal gas law equation: PV = nRT. Let's break it down.

First, we need to convert the temperature from Celsius to Kelvin by adding 273.15. So, 27 degrees C + 273.15 = 300.15 K.

Next, we need to convert the pressure from mmHg to atm by dividing it by 760. So, 900 mmHg / 760 mmHg/atm = 1.184 atm.

Now, we can rearrange the ideal gas law equation to solve for mass, which is given by the equation: m = (molar mass) * (n), where n = PV/RT.

The molar mass of NH3 is 17.03 g/mol.

Substituting in the values, we get:
n = (1.184 atm) * (15.0 L) / [(0.0821 atm L/mol K) * (300.15 K)]
n ≈ 0.8473 mol

Now, multiply the molar mass by n to get the mass of NH3:
m = (17.03 g/mol) * (0.8473 mol)
m ≈ 14.44 g

So, the mass of 15.0 L of NH3 at 27 degrees C and 900 mmHg is approximately 14.44 grams.

To calculate the mass of a gas, we need to use the ideal gas law equation, which relates the pressure, volume, temperature, and number of moles of a gas.

The ideal gas law equation is:

PV = nRT

Where:
P = pressure of the gas (in atmospheres)
V = volume of the gas (in liters)
n = number of moles of gas
R = the ideal gas constant (0.0821 L * atm/mol * K)
T = temperature of the gas (in Kelvin)

First, let's convert the given values to the correct units for the ideal gas law equation.

Temperature:
The given temperature is 27 degrees Celsius. We need to convert it to Kelvin by adding 273.15.
T = 27 + 273.15 = 300.15 K

Volume:
The given volume is 15.0 L, which is already in the correct units.

Pressure:
The given pressure is 900 mmHg. We need to convert it to atmospheres by dividing by the conversion factor:
1 atm = 760 mmHg
P = 900 mmHg / 760 mmHg = 1.18 atm

Now that we have the values in the correct units, we can rearrange the ideal gas law equation to solve for the number of moles (n) of NH3.

n = PV / RT

Substituting the values into the equation:

n = (1.18 atm) * (15.0 L) / (0.0821 L * atm/mol * K * 300.15 K)

Calculating this will give us the number of moles of NH3.

n ≈ 0.732 moles

Finally, to find the mass of NH3, we can use the molar mass of ammonia (NH3), which is approximately 17.03 g/mol.

Mass of NH3 = n * molar mass
Mass of NH3 ≈ 0.732 moles * 17.03 g/mol

Calculating this will give us the mass of NH3.

Mass of NH3 ≈ 12.2 g