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# Chemistry equations

Thu 31 Jan, 2008 07:27 pm
I need help with the following problems because I do not understand how to do them. I appreciate your help and thanks once again.

7. in 2CO(grams) + O2(grams) => 2CO2(grams), what is the ratio of moles of oxygen used to moles of CO2?

9. when iron rusts in air, iron(III) oxide is produced. how many moles react with 2.4 iron in the rusting reaction? 4Fe(s) + 3O2(g) => 2Fe2O3(s)

18. iron (III) oxide is formed when iron combines with oxygen in the air. how many grams of Fe2O3 are formed when 16.7g of Fe reacts completely with oxygen? 4Fe(s) + 3O2(grams) => 2Fe2O3(s)

24. aluminum reacts with sulfuric acid to produce aluminum sulfate and hydrogen gas. how many grams of aluminum sulfate would be formed if 250g H2SO4 reacted with aluminum?

25. how many liters of chlorine gas can be produced when .98L of HCl react with excess O at STP? 4HCl(g) + O2(g) => 2Cl(g) + 2H2O(g)

31. which would be the limiting reagent in the reaction: 2H2(g) + O2(g) => 2H2O(g)

a. 50 molecules of H2
b. 50 molecules of O2

34. at STP, how many liters of oxygen are required to react completely with 3.6 liters of hydrogen to form water? 2H(g)+ O2(g) => 2H2O(g)

35. if 8.6L of H2 reacted with 4.3L of O2 at STP, what is the volume of gaseous water collected (assuming none of it condenses)? 2H2(g) + O2(g) => 2H2O(g)

38. how many liters of O2 are needed to react completely with 45L of H2S at STP? 2H2S(g) + 3O2(g) => 2SO2(g) + 2H2O(g)

41. for the reaction 2Na(s) + Cl2(g) => 2NaCl(s), how many grams of NaCl could be produced from 103g of Na and 13L of Cl2 (at STP)?

43. how many grams of CO are needed to react with an excess of Fe2O3 to produce 209.7gFe? Fe2O3(s) + 3CO(g) => 3CO2(g) + 2Fe(s)
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Ragman

1
Thu 31 Jan, 2008 07:45 pm
why not do your own homework? What will you do when you need answers get into your career?
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waltz

1
Thu 31 Jan, 2008 07:50 pm
because I need help?

what will I do when I need answers to get into my career? good question. maybe I will ask for answers on here?
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Jim

1
Thu 31 Jan, 2008 08:22 pm
Waltz - this really isn't hard, but you do have to try.

Let's look at number 31:

31. which would be the limiting reagent in the reaction: 2H2(g) + O2(g) => 2H2O(g)

a. 50 molecules of H2
b. 50 molecules of O2

You are reacting hydrogen (H2) plus oxygen (O2) to make water. This is about as easy as it gets. The key to the problem are the ratios of how much hydrogen is used against how much oxygen is used. In this case, the ratio is 2:1. You use 2 molecules of hydrogen to one molecule of oxygen to make the water. That's 2:1. The problem statement says you start with an equal amount of hydrogen (50 molecules) and oxygen (50 molecules). Since you are using the hydrogen up twice as fast as the oxygen, the hydrogen will run out first. That's what "limiting reagent" means - what will run out first.

OK, let's look at number 7:

7. in 2CO(grams) + O2(grams) => 2CO2(grams), what is the ratio of moles of oxygen used to moles of CO2?

The numbers given in the equation are:

2 of CO (a reactant, left side of the equal sign)
1 of O2 (a reactant, left side of the equal sign)
2 of CO2 (a product, right side of the equal sign)

so, the ratio of oxygen (that's the O2) to the carbon dioxide (that's the CO2) is 1:2.

Now it's not going to do you any good for us to do your homework for you. You have to learn how to figure it out for yourself. Why don't you pick two or three of these problems, AND TRY. After you've spun your wheels for awhile, then post specifically what you have done, and specifically what you don't understand, and I, or some other A2Ker will be happy to give you a nudge in the right direction.
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waltz

1
Thu 31 Jan, 2008 09:25 pm
9. when iron rusts in air, iron(III) oxide is produced. how many moles of oxygen react with 2.4 iron in the rusting reaction? 4Fe(s) + 3O2(g) => 2Fe2O3(s)

a) 4Fe(s) + 3O2(g) ==> 2Fe2O3(s)

b) convert everything to moles:
Fe = 2.4/55.8

which is 0.0624 mols.

c) moles oxygen: .0624 times (3 moles O2/4 moles Fe)

=0.0468 O2 moles

did i do number 9 correctly?
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waltz

1
Thu 31 Jan, 2008 10:36 pm
this is the only one I don't understand:

41. for the reaction 2Na(s) + Cl2(g) => 2NaCl(s), how many grams of NaCl could be produced from 103g of Na and 13L of Cl2 (at STP/standard temperature pressure)?
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raprap

1
Thu 31 Jan, 2008 10:49 pm
Your still in limiting reactants. Use the mol wt of Na to determine how many moles of Na you have and the Ideal gas law (PV=nRT) to determine how many moles of Cl2 you have---although as a shortcut you can use the relation that 1 mole of any ideal gas at STP is a standard volume (I think it is 22.4L).

Rap
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Jim

1
Fri 1 Feb, 2008 07:05 pm
For Number 41:

For every one Chlorine molecule (Cl2) that gets consumed (used) in the reaction, two Sodium atoms (Na) get consumed (used).

The first thing to do is to calculate how many moles of both reactants you have. This is done differently for the sodium (because you are given the weight of the sodium), and for the chlorine (because you are given the standard volume of the Cl).

Here's the bad news - I work in the American Engineering System of units, where as your school works in SI. Sorry. Life's a bitch.

For sodium, the molecular weight is 22.98. 103g/454 = 0.2269 pounds of sodium. 0.2269/22.98 = 0.00987 pound moles Na.

For the chlorine, we'll use the Ideal Gas Law PV=nRT. Applying a little algebra to solve for N = PV/RT. P= absolute pressure, v=volume (given in the problem statement), R=a constant = 10.73, and T = absolute temperature, and n = the number of moles of the gas.

n = (14.7 psia) * (0.459 ft3) / (10.73) * (492 R) = 0.001278 pound moles Cl2.

Where did I get these numbers? 13 liters/28.31685 = 0.459 cubic feet.

10.73 is the ideal gas constant in the American Engineering system of units.

492 R is the absolute temperature. It is 460 degrees added to the temperature in F. Since STP is at 32 F, the absolute temperature 32+460=492 R.

So, you have 0.00987 pound moles Na, and 0.001278 pound moles Cl2. Remembering the 2:1 ratio they get consumed (used up) at tells you that the Cl2 gets used up first. Or:

all 0.001278 pound moles Cl2 is used

2*0.001278 = 0.002556 pound moles Na gets used

0.00987-0.002556 = 0.007314 pound moles of Na is not used.

0.002556 pound moles of NaCl is produced.

The molecular weight of NaCl is 22.98+35.453 = 58.433.

0.002556 lbmoles * 58.433 = 0.1494 pounds NaCl

*454 = 67.8 grams NaCl.
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Jim

1
Fri 1 Feb, 2008 07:16 pm
For number 9:

dividing 2.4 grams of Iron (Fe) by the molecular weight of Iron (55.8):

2.4 / 55.8 = 0.043 gram moles Iron.

The Iron and the Oxygen are consumed in a 4 : 3 ratio. This means that for every four atoms of Iron that are used, three molecules of Oxygen are used.

0.043 * 3/4 = 0.0323 gram moles of Oxygen are used.

And if you wanted to know the weight of the oxygen used, you would multiply the 0.0323 gram moles by the molecular weight of diatomic (that means there are two of them) oxygen 31.999.

Heck, it's the weekend now. That's enough chemisery for me.
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