Stoichiometry question

[joonkimdds]

---

Members don't see this ad.

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.