Chelsea Q.

# Honors Chemistry Help

1. Phosphorous combines in several different forms. The density of a phosphorous vapor at 310 celsius and 101.3kPa is 2.64g/L. What is the molecular formula of phosphorous?
2. A 2.650g sample of a gas occupies a volume of 428ml at 98.9kPa and 24.3 celsius. Analysis of this compound shows it to be 15.5% Carbon, 23.0% Cl, and 61.5% Fluorine. What is the molecular formula of this compound.
3. Suppose 105L of NH3 (g) and 285L of O2 (g) are allowed to react until this reaction is complete. All volume measurements are made at 200 celsius and 30.4kPa. Which gas, ammonia or oxygen, remains at the end of the reaction? How many moles of it would there be?

Balanced equation: 4NH3 (g) + 7O2 (g) --> 6H2O (g) + 4NO2 (g)

1. An unknown diatomic gas effuses at a rate that is only 0.355 times that of O2 at the same temperature. What is the identity of the unknown gas?

## 3 Answers By Expert Tutors

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Jaimie C.

Suppose 105L of NH3 (g) and 285L of O2 (g) are allowed to react until this reaction is complete. All volume measurements are made at 200 celsius and 30.4kPa. Which gas, ammonia or oxygen, remains at the end of the reaction? How many moles of it would there be? Balanced equation: 4NH3 (g) + 7O2 (g) --> 6H2O (g) + 4NO2 (g) This seems to be a limiting reactant problem. But you can't really start a limiting reactant problem if you don't have moles. Notice you have all the information needed to use the Ideal Gas Equation, and use that equation to fine the moles you need to proceed to determining the limiting reactant. NH3 PV=nRT P=30.4kPa=30400Pa V=105L=0.105m3 n=? R = 8.3145 m3 *Pa *mol-1 *K-1 T=200C=473K 30400Pa*0.105m3=n*8.3145 m3 *Pa *mol-1 *K-1 *473K n=0.812 mol NH3 O2 P=30400Pa V=285L=0.285m3 n=? R = 8.3145 m3 *Pa *mol-1 *K-1 T=473K 30400Pa*0.285m3=n*8.3145 m3 *Pa *mol-1 *K-1 *473K n=2.20 mol O2 Now that we are in moles, we can use a variety of methods to figure out the limiting reactant. I am going to do the mole of reaction method. One mole of reaction is when your starting materials react, based on the coefficients in a balanced equation. 1 mol-rxn = 4 mol NH3 = 7 mol O2 The moles of reaction are directly related to how many times the reaction can occur, so the reactant that has less will be the limiting, as the reaction can occur less times based on it. 0.812 mol NH3 * 1mol-rxn/4molNH3= 0.203 mol-rxn 2.20 mol O2 *1 mol-rxn/7molO2= 0.314 mol-rxn Thus, NH3 is the limiting reactant since the reaction can occur less times with it. Using this and the coefficients of the equation, we can calculate how much oxygen will be leftover after the ammonia reacts to completion. 0.812 mol NH3 *7molO2/4molNH3 = 1.42molO2 used up 2.20-1.42=0.78 mol O2 remaining
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01/21/20

Jaimie C.

An unknown diatomic gas effuses at a rate that is only 0.355 times that of O2 at the same temperature. What is the identity of the unknown gas? This is an effusion problem, so you will need Graham's Law of Effusion rate of effusion A/rate of effusion B=√MB/MA It's hard for me to show with the typing options I have on Wyzant, but the second half of the equation says the square root of the quantity MB divided by MA. These are the molar masses of the two different substances. This equation gives us a way to use the ratio of effusion rates to find the molar mass of the unknown gas. Let's look at what we know to find what we don't know. molar mass of O2= 32.00g/mol molar mass of mystery gas=?? rate of effusion of O2=?? rate of effusion of mystery gas=0.355 times that of O2 Luckily, the rate of effusion for oxygen won't matter much. We can use the fact that the mystery gas is 0.355 times more than it to make our ratio. All ratios are division. For example, if you have 3 sandwiches and I have 5, I can make a ratio of my 5 sandwiches for every one of your 3 as 5/3. Divide that out to 1.6666, and I have 1.6666 sandwiches for every 1 sandwich you have. With the mystery gas, you can think of it in terms of 1, if that helps. For every 1 whatever rate of effusion unit the oxygen does, the mystery gas does 0.355 effusion. Turn it into division to make it a ratio, and you get 0.355/1 which is just 0.355. You can sub this information in as follows: rate of effusion of O2=1 rate of effusion of mystery gas=0.355 This still fits with the "0.355 times as oxygen" statement in the given information. Now fill in the equation and get the molar mass. Let's call the mystery gas substance A and the oxygen substance B to match the formula. rate of effusion A/rate of effusion B=√MB/MA 0.355/1=√32.00g/mol/MA Do the math, and you'll get MA= 254g/mol Remember that this is a diatomic gas. Cut this molar mass in half, then find the gas it matches on the periodic table. 254/2=127g/mol This most closely matches the molar mass of iodine, so your mystery gas is I2.
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01/21/20

Jaimie C.

I tried to write this all very neatly, but the comment system got rid of all the spacing I did. Sorry if it's hard to read!
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01/21/20

Barry M.

tutor
Essentially you got these. However, I suggest just one slight correction. At the very end of the 3rd problem, 2.20 - 1.42 = 0.78, not 0.58. You probably subtracted from 2.00 instead of 2.20. Otherwise, your results are in general agreement with mine.
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01/21/20

Jaimie C.

Oh no! That is a very major typo. Thank you!
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01/21/20

Tutor
5.0 (162)