[magicman]

I know i apologize b4 hand for the double posts, its just as im working im comming across stuff i have yet to encounter examples of.

2 problems which honestly i dont have a clue how to do.

26.11ml of a 0.41M stock solution of CaCl2 is mixed with 101.3ml of water. Calculate the chloride ion concentration in the new solution.




Consider the reaction shown below:

4 NH3 + 5 O2 --> 4 NO + 6 H2O

Calculate the grams of NO which could be produced from the reaction of 43.79g NH3 with 169.44g O2.



When they give me so much numbers i get confused...

<-- Chem n00b who didnt do too hot on his 1st test...  

[mike]

Hello again....

To tackle these problems write down the data that you are given, and convert to SI units:

1.

v1 = 26.11mL = 0.02611L
c1 = 0.41M

v2 = 101.3 + 26.11 = 127.41mL = 0.12741L
c2 = ?

2.

Write down the equation for the reation and balance it. Then write down all the data you are given.

Useful equations:

n = m/M
c = n/v
c1.v1 = c2.v2

n = moles
m = mass
M = molecular weight
c = concentration (c1=intial, c2 = final etc)

[magicman]

For question 1.

For c2 am i to multiply it with v2 ?

I understand how you set it up somewhat (im looking @ my notes..)

.012741L x 0.41 ?

[mike]

rearrange the equation:

c1.v1 = c2.v2

therefore:

c2 = c1.v1/v2

[magicman]

and this would be the new concentration for the ion chloride solution ?

I am just refreshing this untill there is a reply. If you have AIm,MSN or something i would greatly appreciate if you can help me if its not to much trouble for you.

[mike]

Sorry, no instant messaging here for me

c2 would be the concentration of CaCl2 in the new solution.

everyone 1 mole of CaCl2 contains 2 moles of Cl- ions...

[magicman]

ok

so CaCl2

1 mol = 110 g

then divide by 2 so every 1 mol of CaCl2 is = to 2 Cl

so 55 is my final answer ?

55 g ?

[mike]

It depends on what the question wants.

Your original question asks you to work out the chloride ion concentration(?)

So, c2 = c1v1/v2 = 0.084mol.L-1 (this is the concentration of CaCl2 in your new solution) basically this means that in every one litre of solution you have 0.084moles of CaCl2.

However, your question asks for the chloride (Cl-) concentration:

CaCl2 --> Ca2+ + 2Cl-      (when dissolved in water, right?)

So the concentration of Cl- must be twice the concentration of CaCl2 (follow?)

0.168mol.L-1

[magicman]

yes when dissolved in water.

how did you get that .168 M # ?

[mike]

2 x c2  

[magicman]

Mike you have no clue how much you are helping me right now. no Clue....
I greatly appreciate it.




this is where im @ in question 2

gNO = 43.79 g NH3| 1 mol NH3   | 4 mol NO | 20 g NO
                           | 17.03 g NH3| 4 mol NH3| 1 mol NO



If i dont have to use the oxygen why do they give the information? Thats where im left @.

[mike]

Write the equation:

4NH3 + 5O2 --> 4NO + 6H2O

Write down the data you know..

First, work out the limiting reagent by calculating the number of moles of each reactant.

n(NH3) = m/M = 43.79/17 = 2.58 moles
n(O2) = m/M = 169.44/32 = 5.30 moles

so it looks like there is an excess of O2 in this reaction (right?)

hence there is a deficiency of NH3, or in other words the amount of NH3 is going to limit the reaction --> limiting reagent!

Next, the ratio of the limting reagent to the product (NO) is:

NH3 : NO
4:4
1:1

ie for every mole of NH3 that reacts you make a mole of NO (cool?)

So in this reaction you will react 2.58moles of NH3 and make 2.58moles of NO.

To work out the mass of NO we use:

m = n x M = 2.58 x 30 = 77.4g (does this look right?)