# Can you help me about these calculations?

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Hi everyone
I have a chemistry homework but I couldn't solve these questions.Can you help me?

1)As a result of the reaction of 2.08 grams BaCl2 and 3.48 grams K2SO4, 2.2 grams of BaSO4 was obtained. Write the equation for this reaction and calculate the percentage yield. (BaCl2:208.23g/mol ,K2SO4:174.26 g/mol,BaSO4:233.39 g/mol)

2)Calculate the amount of precipitate (AgI) formed when 3 milliliters of 0.1 M AgNO3 reacts with 12 milliliters of 0.1 M Kl and the concentration of ions remaining in the solution. (AgI:234.77 g/mol)

Thanks a lot!

Edited by ms8720
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28 minutes ago, ms8720 said:

Hi everyone
I have a chemistry homework but I couldn't solve these questions.Can you help me?

1)As a result of the reaction of 2.08 grams BaCl2 and 3.48 grams K2SO4, 2.2 grams of BaSO4 was obtained. Write the equation for this reaction and calculate the percentage yield. (BaCl2:208.23g/mol ,K2SO4:174.26 g/mol,BaSO4:233.39 g/mol)

2)Calculate the amount of precipitate (AgI) formed when 3 milliliters of 0.1 M AgNO3 reacts with 12 milliliters of 0.1 M Kl and the concentration of ions remaining in the solution. (AgI:234.77 g/mol)

Thanks a lot!

Since you couldn't solve them you have presumably tried ?

So how far have you got ?

Have you for instance got any proposed reaction equations ?

Edited by studiot
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4 minutes ago, studiot said:

Since you couldn't solve them you have presumably tried ?

So how far have you got ?

Have you for instance calculated any molecular masses ?

Actually I don't know how to solve this. I tried to find out something from other websites but I couldn't find out how to solve. I meant I couldn't find out how to solve them while saying I couldn't solve them.That's my grammar mistake,sorry.

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So does this mean anything to you, if so what ?

$aBaC{l_2}_{\left( {aq} \right)} + b{K_2}S{O_4}_{\left( {aq} \right)} \to nBaS{O_4} \downarrow + mKC{l_{\left( {aq} \right)}}$

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• 1 month later...

Well, I guess it means the OP found an answer from elsewhere faster than you could type.  There is a piece of advice you can give me though.  How can you get those formulae in this textbox (even in the same day)?  Are you dragging and dropping them in from some other software?  What can a simple man like myself use for a bit of mathematical notation?

On 12/9/2020 at 10:53 AM, studiot said:

KCl(aq)

That empty quote is what turns up when I highlight just the last part  KCl(aq).  What is this stuff?     [Late editing:  Wow, it appears when you post it]

Edited by Col Not Colin
Looks different on posting
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10 hours ago, Col Not Colin said:

Well, I guess it means the OP found an answer from elsewhere faster than you could type.  There is a piece of advice you can give me though.  How can you get those formulae in this textbox (even in the same day)?  Are you dragging and dropping them in from some other software?  What can a simple man like myself use for a bit of mathematical notation?

That empty quote is what turns up when I highlight just the last part  KCl(aq).  What is this stuff?     [Late editing:  Wow, it appears when you post it]

How nice it is to have a pleasant converstaion with someone.

Yes I am too old and lazy to learn LaTex or MathML properly so I use a commercial program called MathType.
Here is a screenshot of me setting out a 3 x 3 array

MathType give the option of copy/pasting in various versions of LaTex or other markup languages such as MathML.

I use the MathML output on SF.

Unfortunately MathML is quite expensive.

So here are some sites which offer free online Tex you can copy/[paste from

The only forum I have come across which offered direct LaTex in the editor was AllaboutCircuits dot com, when one of the administrators here at SF was administrator there.

It was a bit 1990s clumsy but could be made to work.

So come on @Dave     and introduce it here.

Final thing,

Due to a glitch in SF software preview doesn't render Tex properly, you also have to refresh your page when viewing Tex initially.

Hope this all helps.

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@studiot  Thank you very much.

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• 2 weeks later...

Since this was originally posted two months ago:

We have aBaCl2(aq)+bK2SO4(aq)nBaSO4+mKCl(aq)

Looking at that I see that there are "a" Ba atoms on the left and "n" Ba atoms on the right so we must have a= n.  I see that there are "2a" Cl atoms on the left and "m" Cl atoms on the right so we must have 2a= m.  Is that there are "2b" K atoms on the left and "m" K atoms on the right so we must have 2b= m.  Finally I see that there are "b" SO4 ions on the left and "n" SO4 ions on the right so we must have b= n.

That is, we have the four equations a= n, 2a= m, 2b= m and b= n.  That is four equations in the four unknowns,  a, b, m, and n.  One difficult is that when the number of equations is equal to the number of unknowns there is typically a unique solution.  But it is obvious that a= b= m= n= 0 is  solution while we want non-zero values!  If there is not a unique solution we can try to solve for all but one unknown in terms of that one.  Let's try to solve for b, m, and n in terms of a.  We have immediately that n= a and m= 2a.  2b= m= 2a so b= a and n= b= a.

So if we arbitrarily take a= 1, b= 1, m= 2, and n= 1.  We have

BaCl2(aq)+K2SO4(aq)→ BaSO4+2KCl(aq)

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6 minutes ago, HallsofIvy said:

Since this was originally posted two months ago:

We have aBaCl2(aq)+bK2SO4(aq)nBaSO4+mKCl(aq)

Looking at that I see that there are "a" Ba atoms on the left and "n" Ba atoms on the right so we must have a= n.  I see that there are "2a" Cl atoms on the left and "m" Cl atoms on the right so we must have 2a= m.  Is that there are "2b" K atoms on the left and "m" K atoms on the right so we must have 2b= m.  Finally I see that there are "b" SO4 ions on the left and "n" SO4 ions on the right so we must have b= n.

That is, we have the four equations a= n, 2a= m, 2b= m and b= n.  That is four equations in the four unknowns,  a, b, m, and n.  One difficult is that when the number of equations is equal to the number of unknowns there is typically a unique solution.  But it is obvious that a= b= m= n= 0 is  solution while we want non-zero values!  If there is not a unique solution we can try to solve for all but one unknown in terms of that one.  Let's try to solve for b, m, and n in terms of a.  We have immediately that n= a and m= 2a.  2b= m= 2a so b= a and n= b= a.

So if we arbitrarily take a= 1, b= 1, m= 2, and n= 1.  We have

BaCl2(aq)+K2SO4(aq)→ BaSO4+2KCl(aq)

Thank you for finishing my calculation.  +1

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