How many grams of carbon dioxide, CO 2 , are present in a 0.150 L flask recorded at 525 mmHg and 32 ºC?

How many grams of carbon dioxide, c02, are present in a 0.150 liter flask recorded at 525 millimeters of mercury and 32 degrees Celsius. So realize here that they're asking us to find grams. But before you can know grams, you need to know the moles of carbon dioxide. So really what they're asking you to find first is the moles of this particular gas. Here we have our volume, we have our pressure, we have our temperature. Using the ideal gas law formula, p v equals n r t, we know that we need to isolate our moles. So divide both sides here by r t. When we do that, we're gonna see that moles equals pressure times volume over r times t. So let's plug in the unit that we know. Our pressure is in millimeters of mercury, but remember, we want the units to be in atmospheres. So we're gonna just gonna do a simple conversion here. One atmosphere is 760 millimeters of mercury. So when we do that, we're gonna get our atmospheres as 0.690789 atmospheres. So take that and plug it in. Next, our volume is already in liters, so plug that in. R is our gas constant, so that's our given number. Remember the units, liters times atmospheres over moles times kelvin. Temperature again has to be in kelvin, so add 273.15 to this value here. And when we do that we're gonna get our our temperature as 305.15 Kelvin. So take that and plug it in. Alright. Now look and see how the units cancel out. Atmospheres cancel out with atmospheres, liters cancel out with liters, kelvins cancel out with kelvins, leaving us with moles of carbon dioxide at the end. So when we do that, we're gonna get initially 4.138 times 10 to the negative 3 moles of carbon dioxide. We don't want moles, we want grams, so one more conversion. We have 1 mole of carbon dioxide, which is made up of 1 carbon and 4 oxygens. When you get their masses from the periodic table, we see that the combined mass of 1 carbon and 2 oxygens is 44.01 grams. Moles cancel out and we're gonna get 1.8 grams of c o 2. It's 0.18 grams because 32 has 2 sig figs, this has 3 sig figs, this has 3 sig figs. Remember, go with the least number of significant figures. So here our answer would be 0.18 grams of carbon dioxide.

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7. Gases

The Ideal Gas Law

General Chemistry

7. Gases

The Ideal Gas Law

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Multiple Choice

How many grams of carbon dioxide, CO₂, are present in a 0.150 L flask recorded at 525 mmHg and 32 ºC?

1.77 g

0.93 g

0.66 g

0.18 g

0.052 g

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Verified step by step guidance

First, convert the temperature from degrees Celsius to Kelvin by adding 273.15 to the Celsius temperature: T(K) = 32 + 273.15.

Next, convert the pressure from mmHg to atm using the conversion factor: 1 atm = 760 mmHg. So, P(atm) = 525 mmHg / 760 mmHg/atm.

Use the Ideal Gas Law equation, PV = nRT, to solve for the number of moles (n) of CO2. Here, P is the pressure in atm, V is the volume in liters, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.

Rearrange the Ideal Gas Law equation to solve for n: n = PV / RT. Substitute the values for P, V, R, and T into the equation to find the number of moles of CO2.

Finally, convert the moles of CO2 to grams using the molar mass of CO2 (approximately 44.01 g/mol). Multiply the number of moles by the molar mass to find the mass in grams: mass (g) = n * 44.01 g/mol.