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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: RustedGuy on February 05, 2019, 11:08:07 AM

Title: Stoichiometry questions about volume and original amount
Post by: RustedGuy on February 05, 2019, 11:08:07 AM
(https://i.stack.imgur.com/VjWym.png)

Question #1: What volume of air is needed to digest 1.00 mol of C6H12O6? Answer in liters.

Question #2: A mixture of KBr and KCl with mass 3.595 grams. It is heated in the presence of Cl2 gas. The mixture converts completely to KCl with mass 3.129 grams. What percentage of the original mixture is KBr?
Title: Re: Stoichiometry questions about volume and original amount
Post by: chenbeier on February 05, 2019, 11:44:39 AM
What ideas do you have to solve?
Title: Re: Stoichiometry questions about volume and original amount
Post by: RustedGuy on February 05, 2019, 11:48:11 AM
I did some research and I determined the hints and the formulas I think I may need to solve these problems. Am I on the right track or am I just creating additional work for myself?
Title: Re: Stoichiometry questions about volume and original amount
Post by: chenbeier on February 05, 2019, 12:04:08 PM
I1. What is chemical equation for the digestion of glucose?
2. If you know that, how much mole Oxygen correspond to 1 mol glucose?
3. What is volume of it
4.  How much air you need.

I Check the mass difference. It corresponds to the gain of Chloride and lost of Bromide.


Title: Re: Stoichiometry questions about volume and original amount
Post by: RustedGuy on February 05, 2019, 02:32:04 PM
I1. Digestion chemical formula for glucose = C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O

2. Is the equation to solve this: 1 mol C6H12O6 * 6 mol O2/1 mol C6H12O6 * 15.999 g O2/1mol = 95.994 g O2 -> 32.0 mol of O2

3. I don't understand this part. What is the volume of what?

4. I don't understand this either. How much air of what do I need?

---

I2. KBr molar mass = 119.002 g/mol
KCl molar mass = 74.5513 g/mol
Cl2 molar mass = 70.906 g/mol
Title: Re: Stoichiometry questions about volume and original amount
Post by: chenbeier on February 05, 2019, 02:50:13 PM
I don`t know what are you calculating.

The formula is correct.

So if you digest 1 mol glucose you need 6 mol O

6 mol O2 correspond to how much liter. Hint Molar volume.

This volume has to be increased to volume of air.

The second question you didn`t solve anything.

Zou have a weight of the mixture KCl/KBr and a weight of KCl. The difference of this weight is resulting by loosing Bromine and gaining Chlorine.

So calculete the difference mole= delta Mass / delta molar mass chlorine-bromine
continue with this
Title: Re: Stoichiometry questions about volume and original amount
Post by: AWK on February 05, 2019, 03:08:48 PM
Question 1
This is a very easy problem. If temperature and pressure are not given use  STP. Chemical engineers use a very good approximation for air content: O2+4N2 (in moles or volumes). You calculated correctly moles of oxygen needed. Having known volume of the ideal gas at STP you can calculate the needed volume of air even without any calculator within less then 2% error.
Title: Re: Stoichiometry questions about volume and original amount
Post by: RustedGuy on February 05, 2019, 03:13:23 PM
I1. 6 mol O2 * 22.4 L O2/1 mol O2 = 134.4 L O2

I2. (3.595-3.129)/(79.904-35.453) = 0.01048
Title: Re: Stoichiometry questions about volume and original amount
Post by: AWK on February 05, 2019, 03:21:27 PM
I 1.
This is a volume of oxygen. Air (O2+4N2) - X times more.
I 2.
Multiply this number by MW KBr and calculate % of KBr in this mixture.
Title: Re: Stoichiometry questions about volume and original amount
Post by: RustedGuy on February 05, 2019, 03:28:09 PM
I1. Can you explain this more please? I don't understand what I do with the calculated number I got. Am I not supposed to multiply by 22.4 L?

I2.

0.01048 * 119.002 g/mol = 1.247
1.247/3.595 * 100 = 34.69%
Is this the final answer to question #2?
Title: Re: Stoichiometry questions about volume and original amount
Post by: AWK on February 05, 2019, 03:38:20 PM
I1
Volume of 1 L oxygen and 4 L of nitrogen equals to ???
I2
perfectly done
Title: Re: Stoichiometry questions about volume and original amount
Post by: RustedGuy on February 05, 2019, 03:42:37 PM
I1. One mole of an ideal gas will occupy a volume of 22.4 liters at STP. 1L is 1/22.4 * 32 = 1.429g for oxygen, 4L is 1/22.4 * 14 = 0.625g for nitrogen

I2. Thank you AWK and chenbeier for your help.
Title: Re: Stoichiometry questions about volume and original amount
Post by: AWK on February 05, 2019, 03:46:35 PM
And you found factor (5) that multiply a volume of oxygen to get the volume of air.
Title: Re: Stoichiometry questions about volume and original amount
Post by: RustedGuy on February 05, 2019, 03:49:31 PM
I1. 5 * 134.4 L of O2 = 672 L of air
Is this the final answer for question #1?
Title: Re: Stoichiometry questions about volume and original amount
Post by: AWK on February 05, 2019, 03:59:47 PM
OK
Title: Re: Stoichiometry questions about volume and original amount
Post by: chenbeier on February 06, 2019, 05:10:33 AM
Quote from: RustedGuy on February 05, 2019, 03:13:23 PM

I2. (3.595-3.129)/(79.904-35.453) = 0.01048

My question is we add Cl2 and removed Br2

Should it (3.595-3.129)/(159,8-70,9) = 0.00524 ?
Title: Re: Stoichiometry questions about volume and original amount
Post by: AWK on February 06, 2019, 05:50:11 AM
in KBr ----- Br (=1/2 Br2) is replaced by Cl (1/2 Cl2)
but for example in CaBr2 your reasoning would be OK.
Title: Re: Stoichiometry questions about volume and original amount
Post by: chenbeier on February 06, 2019, 05:58:37 AM
ok