GC question

[Jul18]

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When we're converting from g---> to atom we gotta use this


g--->mol---->molecule------>atom...is that right?

1) How many moles of Chloride ions are in 0.0750g of Alumninum Chloride?

0.0750g AlCl3 x 1mol/133.5g x 6.02x10^23 mole x 3 Cl- ions/mol

= 1.01 x 10^21 ions Cl?

is that right?

I am really confuse on when to use the Avogadro's number in calculating the ions, atoms, etc... please help

 

[osimsDDS]

When we're converting from g---> to atom we gotta use this


g--->mol---->molecule------>atom...is that right?

1) How many moles of Chloride ions are in 0.0750g of Alumninum Chloride?

0.0750g AlCl3 x 1mol/133.5g x 6.02x10^23 mole x 3 Cl- ions/mol

= 1.01 x 10^21 ions Cl?

is that right?

I think what you want to do is find the moles of Cl- produced, what you have right now is just AlCl3....so first write out the dissociation of AlCl3 and balance

AlCl3 ----> Al^3+ + 3Cl-

Now you can start doing stoicheometry:

.075g AlCl3 x 1mol/133.5g AlCl3 x 3mol Cl-/1mol AlCl3 x 6.02x10^23/1mol = 1.01 x 10^21

You got the same answer but in more difficult problems you have to know how many moles of each element your dealing with...but yea you have the right idea...

 

[DentalDeac]

I dont understand why you guys are using Avagadro's number, its asking for the moles of Cl ions so wouldn't it just be

.075 g AlCl3 X 1mol/133.5g AlCl3 x 3 mol Cl- / 1 mol AlCl3 ?

 

[osimsDDS]

I dont understand why you guys are using Avagadro's number, its asking for the moles of Cl ions so wouldn't it just be

.075 g AlCl3 X 1mol/133.5g AlCl3 x 3 mol Cl- / 1 mol AlCl3 ?

hahah i didnt see that...i thought it was asking for how many atoms...my bad yea its just moles, cut off the avogadro's number from the end of the stoiceometry and you got your answer...

 

[Jul18]

ahhh..my bad..u guys are awesome... thanks!