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December 22, 2014

December 22, 2014

Posted by **Trish** on Wednesday, December 13, 2006 at 9:44pm.

Also, I really have tried these...believe me. I just don't even know where to start..

1. 2KClO3-->2KCl+3O2

Question: What mass of KClO3 must be used in order to generate 5 Liters of O2, measured in STP?

2. NaCl+H2O+CO2+NH3-->NaHCO3+NH4Cl

Question 1: What mass of NaHCO3 can be formed when NH3 and CO2 react with NaCl?

Question 2: If this reaction forms 46 kg of NaHCO3, what volume of NH3 in STP reacted?

Question 3: What volume of CO2 at 5.50 atm and 42 degrees Celcius will be needed to make 100 kg of NaHCO3?

3. 2C4H10 + 13O2-->8CO2 + 10H2O

Q1: If 4.74 grams of butane react with excess oxygen, what volume of CO2 at 150 Degrees celcius and 1.14 atm will be made?

Q2: What volume of O2 at .980 atm and 75 degrees celcius will be consumed by the complete combustion of .500 grams of butane?

Q3: A butane torch has a mass of 876.2 grams. After burning awhile, it has a mass of 859.3 grams. What volume of CO2 at STP was formed while the torch burned?

Q4: What mass of H2O is produced when butane burns and produces 3720 liters of CO2, measured at 35 degrees celcius and .993 atm of pressure?

I WISH I CAN HELP YOU LIKE YOU HELPED ME BUT I'M NOT THAT ADVANCE.I REPLYED TO YOU UNDER MY MESSAGE

Thank you once again :) I hope your feeling better and you recive help with your math.

1. 2KClO3-->2KCl+3O2

Question: What mass of KClO3 must be used in order to generate 5 Liters of O2, measured in STP?

Step 2. Convert what you are given to mols. Remember mols = grams/molar mass for solids OR(for gases) mols = volume in L/22.4.

In this case, mols O2 = 5L/22.4 = 0.2232 mols.

Step 3. Using the balanced equation from step 1, convert mols of what you have (in this case mols oxygen) to mols of what you want (in this case mols KClO3). Use the coefficients in the balanced equation.

mols KClO3 = 0.2232 mols O2 x (2 mol KClO3/3 mols O2) = 0.1488 mols KClO3.

Step 4. Now convert what you have from step 3 (mols KClO3) to grams KClO3. Just redo the appropriate equation from step 2. Since mols = grams/molar mass, then grams = mols x molar mass and grams KClO3 = 0.1488 mol KClO3 x 122.6 g KClO3/1 mol KClO3 = 18.2 g KClO3.

That's really all there is to it. I'll let you practice on the others but you need to learn how to do these by yourself. I hope this helps you. I shall be happy to check the answers for you if you post them but it would be helpful if you showed your work so that I can see how you are solving them.

So I tried working out this problem, but I got confused at step 4....how did you get the 122.6?

That's the molar mass of KClO3 but you need to add up the atomic masses of K, Cl, and 3 Oxygens to see if that is right.

Yessss, I got that exactly by rounded the masses to the tenths. Thank You!

Let me try Number2 Q1 then...

2. NaCl+H2O+CO2+NH3-->NaHCO3+NH4Cl

Question 1: What mass of NaHCO3 can be formed when NH3 and CO2 react with NaCl?

I'm having trouble balancing this...I can't figure it because the H's keep coming up uneven..argggg.

NaCl+H2O+CO2+NH3-->NaHCO3+NH4Cl

1 Na left and right.

1 Cl left and right.

5 H left and 5 right.

3 O left and 3 right.

1 C left and right.

1 N left and right.

Looks good to me.

I don't see any numbers for the amount of NH3 and CO2. 1 mol NaHCO3 (84 grams) will be formed if 1 mol CO2 reacts with 1 mol NH3. But that's the best that can be done if the problem lists no numbers.

Oh wow...I'm retarded...sorry..my mom's making me eat dinner, but I'll be back later. Thank you for your help. :)

I may be gone when you finish but just stay with it. I worked out q2 for problem 2 and obtained 12.2 L NH3 but check my work. For q3 of problem 2, I found 5,594 L. Again, check my work. Post here if you have trouble and someone will help.

omfg i cant even figure that out. so yea. well i need help to with that. lol

:]]

Make a repost at the beginning and tell us what you don't understand about what.

sige

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**Confuzzled**, Saturday, July 25, 2009 at 6:10amWhat are the products of this reaction:

NaHCO3 + NH3 --> H2O + ...

Thanks.

(I assume water is produced given sodium bicarbonate is amphoteric. But basically I need to otherside of the equation.)

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