Quick chemistry problem (practice question).

[Coldkilla]

What is the experimental yield (in grams) when the percent yield is 11.01 % when 84.4 g of barium nitrate reacts in solution with excess sodium phosphate?

I have done this a hundred times and can't seem to figure it out. I know the formula is:

% Yield = Actual Mass/Theoretical Mass X100%

But simply plugging in the numbers in this equation can't be the only thing that is required.

When I create the equation:
Ba(NO3)2 + Na3+PO4

I then balance it:
3Ba(NO3)2 + 2Na3(PO4) --> 6Na(NO3) + Ba3(PO4)2

I don't know where to begin with the given 84.4gBa(NO3)2. I could start an equation like:
84.4gBa(NO3)2 * (1mol Ba(NO3)2 / 261.32g Ba(NO3)2)

to get its molar mass but I have absolutely no idea where to go from here. Any help appreciated.
====================================

Old/previous ANSWERED question:
What is mass (in g) of copper can be recovered from 57.4 mL of a 0.669 M solution of copper(I) chloride and excess hydrogen?

Wouldn't it be 9.0x10^23?

Because this is what I would do:
1. Create the equation: CuCL + H2 --> HCL + Cu
2. Balance it: 2CuCL + H2 --> 2HCL + 2Cu
3. Start the problem:

57.4mL(CuCL) * (1L(CuCL)/1000mL(CuCL)) * (1mol(CuCL)/1L(CuCL)) * (1mol(Cu)/0.669mol(CuCL)) * (6.022x10^23g(Cu)/1mol(Cu)) = 9.0x10^23 grams of Cu.

Am I working on this correctly? If not, where am I going wrong? Thanks.

 

[WraithETC]

By your balanced equation: mols of CuCl = mols of Cu because its the limiting reactant and the coefficients are the same.

Take mols of CuCl and multiply by Copper molar mass to get grams of copper.

Anyways that seems like a shit load of copper for 57.4ml of stuff.

 

[Aaviel]

Originally posted by: Coldkilla
What is mass (in g) of copper can be recovered from 57.4 mL of a 0.669 M solution of copper(I) chloride and excess hydrogen?

Wouldn't it be 9.0x10^23?

Because this is what I would do:
1. Create the equation: CuCL + H2 --> HCL + Cu
2. Balance it: 2CuCL + H2 --> 2HCL + 2Cu
3. Start the problem:

57.4mL(CuCL) * (1L(CuCL)/1000mL(CuCL)) * (1mol(CuCL)/1L(CuCL)) * (1mol(Cu)/0.669mol(CuCL)) * (6.022x10^23g(Cu)/1mol(Cu))= 9.0x10^23 grams of Cu.

Am I working on this correctly? If not, where am I going wrong? Thanks.

should be 63.54 g cu / mol cu

there are 6.022x10^23 molecules in one mol of cu not grams

 

[Coldkilla]

Oh ok so if I may:

1. The balanced equation: 2CuCL + H2 --> 2HCL + 2Cu
2. My reworked equation (the longhand way):
57.4mL(CuCL) * (1L(CuCL)/1000mL(CuCL)) * (0.669mol(CuCL)/1L(CuCL)) * (2mol(Cu)/2mol(CuCL)) * (63.55g(Cu)/1mol(Cu)) = 2.44g(Cu)

or the stacked way:

57.4mL CuCL x

(1L CuCL) /
1000mL CuCL) x

(0.669mol CuCL) /
1LCuCL) x

(2mol Cu) /
2mol CuCL) x

(63.55g Cu) /
1molCu) = 2.44g(Cu)

Edit: Well I keep screwing up then.. Idk how to set this up. I need to do it the long way as I'm a little slow in the head.

 

[WraithETC]

(57.4ml /1000) L * .669 mols/Liter * (2molsCu/2molsCuCl) * 63.55 g/mol = 2.44g

All units cancel expect for grams so this seems right.

Should be right.

 

[Coldkilla]

Alright this seems to make sense. Thanks a bunch!

 

[Coldkilla]

I've only got 1 more question if anyone could help me out. Thanks. (Just added it):

What is the experimental yield (in grams) when the percent yield is 11.01 % when 84.4 g of barium nitrate reacts in solution with excess sodium phosphate?

I have done this a hundred times and can't seem to figure it out. I know the formula is:

% Yield = Actual Mass/Theoretical Mass X100%

But simply plugging in the numbers in this equation can't be the only thing that is required.

When I create the equation:
Ba(NO3)2 + Na3+PO4

I then balance it:
3Ba(NO3)2 + 2Na3(PO4) --> 6Na(NO3) + Ba3(PO4)2

I don't know where to begin with the given 84.4gBa(NO3)2. I could start an equation like:
84.4gBa(NO3)2 * (1mol Ba(NO3)2 / 261.32g Ba(NO3)2)

to get its molar mass but I have absolutely no idea where to go from here. Any help appreciated.

 

[Fenixgoon]

rule #1 - convert to moles. then just figure out what your theoretical yield is, multiply by 0.1101, and convert to grams.

 

[Coldkilla]

Yea I started out thinking I needed moles. So I figured out that from the given 84.4g, there are .32 moles barium nitrate there. I don't understand how I can go from moles of barium nitrate to this "theoretical yield".

I mean... I have moles of barium nitrate, but what am I supposed to compare it to?
2 moles of Na3(PO4)?
6 moles of Na(NO3)?
1 mole of Ba3(PO4)2?

On top of that I don't have another mass (besides the initial 84.4) to work with the next reactant, Na3(PO4).. arg I'm so confused.

 

[nonameo]

I must be taking sub-par chem classes. I haven't seen this in chem I or chem II

 

[Coldkilla]

Well I think I figured it out. But I still had a question regarding this.

What I did was:
1. Took the given 84.4g barium nitrate and figured out that it was .32 moles by deviding its molar mass.
2. I then chose (at random), one of the two products - in this case: 6Na(NO3) and compared its moles with the number of moles in the reactant (barium nitrate).
3. I then multiplied by 6Na(NO3)'s molar mass and got 8.86

So matematically it looked like this:

3Ba(NO3)2 + 2Na3(PO4) --> 6Na(NO3) + Ba3(PO4)2

84.4g Ba(NO3)2 x

1mol Ba(NO3)2 /
261.32g Ba(NO3)2 (molar mass) *

1mol Ba3(PO4)2 /
3 mol Ba(NO3)2 (molar ratios) *

601.84g Ba3(PO4)2 (molar mass) /
1 mol Ba3(PO4)2

= 64.60 *
percent yield (82.8 %) * 100% = 8.86 answer

My question is: Was I right to use Ba3(PO4)2 as to compare it to barium nitrate? Or could I have done it with 6NaCL and got the same thing?

 

[Rockinacoustic]

I hated this stuff in Gen Chem. Once I took Orgo lab and saw my yields in real life it finally clicked.

 

[Fenixgoon]

Originally posted by: Coldkilla
Yea I started out thinking I needed moles. So I figured out that from the given 84.4g, there are .32 moles barium nitrate there. I don't understand how I can go from moles of barium nitrate to this "theoretical yield".

I mean... I have moles of barium nitrate, but what am I supposed to compare it to?
2 moles of Na3(PO4)?
6 moles of Na(NO3)?
1 mole of Ba3(PO4)2?

On top of that I don't have another mass (besides the initial 84.4) to work with the next reactant, Na3(PO4).. arg I'm so confused.

"excess barium nitrate" means as much as you need for things to react. in other words, don't worry about it.

write your balanced reaction. calculate how many moles of sodium nitrate and barium phosphate you will (ideally) get.

multiply by 0.1101 to get 11.01% of that yield.

convert to grams.

 

[WraithETC]

Originally posted by: nonameo
I must be taking sub-par chem classes. I haven't seen this in chem I or chem II

The stuff hes doing is like the first chem class in an introductory chem series.

 

[nonameo]

Originally posted by: WraithETC

Originally posted by: nonameo
I must be taking sub-par chem classes. I haven't seen this in chem I or chem II

The stuff hes doing is like the first chem class in an introductory chem series.

Maybe I just don't remember it then.