need help with chemistry question

[anneee]

Hey guys, i need help with the following questions for chemistry

1. what masses of barium sulfate and calcium sulfate must be mixed so that 1000g of a mixture contains equal numbers of moles of each?

and also

2. A mixture of calcium and magesium carbonate of mass 1.84g was strongly heated until no futher loss in mass was detected. The residue had mass of 0.900g. what percentage by mass of the mixture was calcium carbone?

( i also tried using simultaneous equations to work out qs.2 but somehow my answer wasn't right)

thank you and any help would be greatly appreciated (:

[Nobby]

Okay here's my piece

1. x(M(BaSO4)+M(CaSO4))=1000g
 => x(233.3+136.1)=1000
 => x=1000/(233.3+136.1)
        =2.708mol

2. Umm...what?

[HERculina]

Okay here's my piece

1. x(M(BaSO4)+M(CaSO4))=1000g
 => x(233.3+136.1)=1000
 => x=1000/(233.3+136.1)
        =2.708mol

2. Umm...what?

1. After working out no. Of moles needed overall like what Nobby did, you divide 2.708 by 2 = 1.354 mol as their are equal amounts of BaSO4 and CaSO4. And then to work out mass of each i would do: m(BaSO4) = 1.354 * 233.3= 315.9 g and m(CaSO4)= 1.354 * 136.1 = 184.3g.    2. Still trying to work out this next question ^^

[Nobby]

Okay here's my piece

1. x(M(BaSO4)+M(CaSO4))=1000g
 => x(233.3+136.1)=1000
 => x=1000/(233.3+136.1)
        =2.708mol

2. Umm...what?

1. After working out no. Of moles needed overall like what Nobby did, you divide 2.708 by 2 = 1.354 mol as their are equal amounts of BaSO4 and CaSO4. And then to work out mass of each i would do: m(BaSO4) = 1.354 * 233.3= 315.9 g and m(CaSO4)= 1.354 * 136.1 = 184.3g.    2. Still trying to work out this next question ^^

haha forgot to do the entire second part of the question...

[Bhootnike]

Have you got an answer for part 2? J

[Bhootnike]

Regardless, I attempted the question.. :p

Heres my working:

Calcium carbonate is CaCO3, and has molar mass of 100, likewise, magnesium carbonate is MgCO3, and its molar mass is 84.
Assumed the following equations since it is heated...

CaCO3 --> CaO + CO2                  ----1

MgCO3 --> MgO + CO2                   ---2

Now you make some ratios from given information, which you will use to solve the values required. Be sure to label equations for ease!!

m(CaCO3)          100
-----------     =   ------                  -----3
m(CaO)              56

m(MgCO3)          84
-----------     =   ------                    -----4
m(MgO)              40

We thus know that:

m(CaCO3) + m(MgCO3) = 1.84 g        ------5
and
m(CaO) + m(MgO) = 0.9g                 ------6


From equation 3, you can rearrange to find that:

                   100
CaCO3=      ----- X m(CaO)            -------- 7
                    56
Likewise, from eq. 4, you can rearrange to find that:

                          84
m(MgCO3) =      ----- X m(MgO)                      -----8
                          40

Therefore, since we've expressed each of the compounds, MgCO3 and CaCO3 in terms of its respective products, Cao And MgO, you can rewrite equation 5 as:


  100                         84
[ ----- X m(CaO)] + [----- X m(MgO)] = 1.84                              -----9
  56                           40

From eq. 6, you can rearrange to express MgO in terms of CaO, such that:

m(MgO) = 0.9 - m(CaO)

Now, substitute the above equation into equation 9, to get:

  100                         84
[ ----- X m(CaO)] + [----- X 0.9-m(CaO)] = 1.84                               
  56                           40

Let m(Cao) = x, and solve for x.

You do the number crunching and find x to be 0.15909 i think!
therefore m(CaO) = 0.16

To find CaCO3, all you have to do is times by the ratio you get when you divide the two masses. i.e, 1.84 / 0.9 = 2.044444
So m(CaCO3) = 0.32711111111


Now to find percentage by mass,

0.327111
-----------
1.84

=0.1777777777778

Therefore, percentage by mass of CaCO3 = 17.7778 % = 18%

I hope thats correct!!

EDIT!

I'll go back to "You do the number crunching and find x to be 0.15909 i think!
therefore m(CaO) = 0.16"

I think what you're meant to do is... :

From equation 3

100
----  = 1.785714286
56

Use this ratio to determine the mass of CaCO3 , i.e. 1.79 x 0.16 = 0.2864
This is the mass of CaCO3, therefore:

0.2864/1.84 =0.15556521739
Therefore, 15.5556% = 16%

Hopefully that's better?!?

@PanicMode - i dont know honestly which one is right,  hahaha

[Panicmode]

See attached scan for question 2.

Bhootnike, you have the right idea but near the end, you need to use the calculated mass of CaO to find the mass of  CaCO3 (by using mol ratios).

Unless I did something wrong....

[Bhootnike]

Edited my post above

[anneee]

hey all! thanks for all the help, i think i know where i went wrong now! and the answer for part 2 is 13.5%

[Mao]

I'm not a fan of the relative mass approach. Here's how I would go about this:

2. A mixture of calcium and magesium carbonate of mass 1.84g was strongly heated until no futher loss in mass was detected. The residue had mass of 0.900g. what percentage by mass of the mixture was calcium carbone?

We know a few things:
-net mass loss is all loss of CO2.
m(CO2) = 0.94
-each carbonate can lose one CO2.
n(CaCO3) + n(MgCO3) = n(CaO) + n(MgO) = n(CO2)
m(CaCO3) + m(MgCO3) = 1.84
m(CaO) + m(MgO) = 0.9

n(CaO)*56.1 + n(MgO)*40.3 = 0.9
n(CaO) + n(MgO) = 0.94/44

Solve simultaneously to get the mole ratio, hence mass ratio, etc etc.

(just re-read the question, it seems the smarter way to go about it would be to use the carbonate equations. Anyhow, the same method applies)

[Bhootnike]

hey all! thanks for all the help, i think i know where i went wrong now! and the answer for part 2 is 13.5%

well i guess i shouldve done what panicmode said haha