Stoichiometry question

[joonkimdds]

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This is an example from Kaplan blue book page 232.

How many grams of calcium chloride are needed to prepare 72g of silver chloride according to the following equation?
CaCl2 + 2AgNO3 -> Ca(NO3)2 + 2AgCl

solution on the book is
72g AgCl x (1mol of AgCl/144g AgCl) x (1mol CaCl2/2mol AgCl) x (110g CaCl2/1mol CaCl2) = 27.5 g CaCl2

I really don't like this way of explanation. I know that even my Chemistry textbook explains the same way but it's not a good way for me to understand.
so I used my way, and I just don't know why i get different answer.
so could u guys help me finding out what my error is?
below is what I so called "my way" to make it look a lot simpler and easier to understand.

1 mol of CaCl2 for 2 mol of AgCl(144g)
? mol of CaCl2 for 1 mol of AgCl(72g)

and i get ? = 0.5

so 0.5 mol of CaCl2 for 1 mol of AgCl
Since 1 mol of AgCl=72, I get g of 0.5 mol of CaCl2 for the answer.

molar mass of AgCl is 110g.
but then it's 110 divide by 2 = 55 g of CaCl2
I am supposed to have 27.5 g of CaCl2.

so I believe my mistake is, instead of divide 110 by 2, i should divide by 4 to get 27.5 g of CaCl2, so where did i make mistake that make me get 2 instead of 4?

 

[I AM SARA]

This is an example from Kaplan blue book page 232.

How many grams of calcium chloride are needed to prepare 72g of silver chloride according to the following equation?
CaCl2 + 2AgNO3 -> Ca(NO3)2 + 2AgCl

solution on the book is
72g AgCl x (1mol of AgCl/144g AgCl) x (1mol CaCl2/2mol AgCl) x (110g CaCl2/1mol CaCl2) = 27.5 g CaCl2

I really don't like this way of explanation. I know that even my Chemistry textbook explains the same way but it's not a good way for me to understand.
so I used my way, and I just don't know why i get different answer.
so could u guys help me finding out what my error is?
below is what I so called "my way" to make it look a lot simpler and easier to understand.

1 mol of CaCl2 for 2 mol of AgCl(144g)
? mol of CaCl2 for 1 mol of AgCl(72g)

and i get ? = 0.5

so 0.5 mol of CaCl2 for 1 mol of AgCl
Since 1 mol of AgCl=72, I get g of 0.5 mol of CaCl2 for the answer.

molar mass of AgCl is 110g.
but then it's 110 divide by 2 = 55 g of CaCl2
I am supposed to have 27.5 g of CaCl2.

so I believe my mistake is, instead of divide 110 by 2, i should divide by 4 to get 27.5 g of CaCl2, so where did i make mistake that make me get 2 instead of 4?

the way you do it makes 0 sense.

Its just wrong.

 

[joonkimdds]

the way you do it makes 0 sense.

Its just wrong.

why does it make 0 sense?
All I did was using the fact that mol and g always have the same ratio.

For example, if there was a question that says
"Determine the moles of O2 needed to roast 10 mol of Cu2S
in a equation 2Cu2S + 3O2 -> 2Cu2O + 2SO2

then since 3 mole of O2 is needed for every 2 mol of Cu2S
what I do is

3 mol of O2 for 2 mol of Cu2S
? mol of O2 for 10 mol of Cu2S

so ? = 15 mol of O2

the only difference between this 2nd example I made and the 1st example that I posted is that 1st example asks about gram and 2nd example asks about mol.

 

[joonkimdds]

the way you do it makes 0 sense.

Its just wrong.

here is another example that can prove my way can make sense.


"Determine the mass(g) of SO2 formed from 10mol of Cu2S
2Cu2S + 3O2 -> 2Cu2O + 2SO2


2 mol of Cu2O for 2 mol of SO2
10 mol of Cu2O for 10 mol of SO2

since SO2 has 64.07g, 10 mol should have 64.07 times 10 = 640.7g and it's the right answer with the same way I use my way.

 

[Flipper405]

This is an example from Kaplan blue book page 232.

How many grams of calcium chloride are needed to prepare 72g of silver chloride according to the following equation?
CaCl2 + 2AgNO3 -> Ca(NO3)2 + 2AgCl

solution on the book is
72g AgCl x (1mol of AgCl/144g AgCl) x (1mol CaCl2/2mol AgCl) x (110g CaCl2/1mol CaCl2) = 27.5 g CaCl2

I really don't like this way of explanation. I know that even my Chemistry textbook explains the same way but it's not a good way for me to understand.
so I used my way, and I just don't know why i get different answer.
so could u guys help me finding out what my error is?
below is what I so called "my way" to make it look a lot simpler and easier to understand.

1 mol of CaCl2 for 2 mol of AgCl(144g)
? mol of CaCl2 for 1 mol of AgCl(72g)

and i get ? = 0.5

so 0.5 mol of CaCl2 for 1 mol of AgCl
Since 1 mol of AgCl=72, I get g of 0.5 mol of CaCl2 for the answer.

molar mass of AgCl is 110g.
but then it's 110 divide by 2 = 55 g of CaCl2
I am supposed to have 27.5 g of CaCl2.

so I believe my mistake is, instead of divide 110 by 2, i should divide by 4 to get 27.5 g of CaCl2, so where did i make mistake that make me get 2 instead of 4?

I think your mistake is that you're using 144g as 2 mol AgCl when 144g is 1 mol AgCl. Regarding the method, I find the factor-label method a lot easier than the way you are proposing. I guess it's just personal preference...


*edit*

The problem was that you used the wrong molar mass for AgCl, and I think you also assumed in your method that you start with 1 mol AgCl when you actually start with 0.5 mol of AgCl.

So using your method... ( I think)

1 mol CaCl2 for 2 mol AgCl(288g)
0.5 mol CaCl2 for 1 mol AgCl(144g)
? mol CaCl2 for 0.5 mol AgCl(72g)
?=0.25

0.25 mol CaCl2 for 0.5 mol AgCl
0.5 mol AgCl=72, g of 0.25 mol CaCl2 is answer.

molar mass of CaCl2 is ~110 g
110/4=27.5g of CaCl2

Too complicated for my taste
But good luck anyway. 👍

 

[joonkimdds]

I think your mistake is that you're using 144g as 2 mol AgCl when 144g is 1 mol AgCl. Regarding the method, I find the factor-label method a lot easier than the way you are proposing. I guess it's just personal preference...


*edit*

The problem was that you used the wrong molar mass for AgCl, and I think you also assumed in your method that you start with 1 mol AgCl when you actually start with 0.5 mol of AgCl.

So using your method... ( I think)

1 mol CaCl2 for 2 mol AgCl(288g)
0.5 mol CaCl2 for 1 mol AgCl(144g)
? mol CaCl2 for 0.5 mol AgCl(72g)
?=0.25

0.25 mol CaCl2 for 0.5 mol AgCl
0.5 mol AgCl=72, g of 0.25 mol CaCl2 is answer.

molar mass of CaCl2 is ~110 g
110/4=27.5g of CaCl2

Too complicated for my taste
But good luck anyway. 👍

Flipper~ u r my hero~ 😍

 

[I AM SARA]

Flipper~ u r my hero~ 😍

amen.