Strange result in stoichiometry limiting/excess problem.

[pc2-brazil]

Salutations,
The answer obtained below doesn't match any of the options given in the enunciation.
Thank you in advance.

Homework Statement

A sample of dolomitic limestone, containing 60% of calcium carbonate and 21% of magnesium carbonate, goes through decomposition when heated, according to the following equation:

[tex]CaCO_3_{(s)} + MgCO_3_{(s)} \rightarrow CaO_{(s)} + MgO_{(s)} + 2CO_2_{(g)}[/tex]

The mass of calcium oxide and the mass of magnesium oxide, in grams, obtained from the burning of 1 kg of limestone are, respectively:

a) 60; 21.
b) 100; 84.
c) 184; 96.
d) 336; 100.
e) 600; 210.

Homework Equations

The Attempt at a Solution

Molar masses:
CaCO₃: 40 + 12 + 3*16 = 52 + 48 = 100 g.
MgCO₃: 24,5 + 12 + 3*16 = 36,5 + 48 = 84,5 g.
CaO: 40 + 16 = 56 g.
Mass of CaCO₃ and MgCO₃ in 1 kg of dolomitic limestone (1000 g):
Mass of CaCO₃: 60% of 1000 g = 600 g.
Mass of MgCO₃: 21% of 1000 g = 210 g.
Discover what is the limiting reagent:
[tex]1 mol CaCO_3 \rightarrow 1 mol MgCO_3[/tex]
[tex]100 g \rightarrow 84,5 g[/tex]
[tex]600 g \rightarrow x[/tex]
x = 84,5 * 6 = 507 g of MgCO₃ for consuming all of the CaCO₃; since we only have 210 g:
MgCO₃ => limiting.
Use the mass of MgCO₃ to discover how many CaO will be produced:
[tex]1 mol MgCO_3 \rightarrow 1 mol CaO[/tex]
[tex]84,5 g \rightarrow 56 g[/tex]
[tex]210 g \rightarrow y[/tex]
[tex]y = \frac{210 \times 56}{84,5}[/tex]
y = 139,17 g.
But this mass is not in the answer options. According to the book, the right answer is D (336 g of CaO and 100 g of MgO). What is wrong, then?
Thank you in advance.

 

[GCT]

Salutations,
The answer obtained below doesn't match any of the options given in the enunciation.
Thank you in advance.

Homework Statement

A sample of dolomitic limestone, containing 60% of calcium carbonate and 21% of magnesium carbonate, goes through decomposition when heated, according to the following equation:

[tex]CaCO_3_{(s)} + MgCO_3_{(s)} \rightarrow CaO_{(s)} + MgO_{(s)} + 2CO_2_{(g)}[/tex]

The mass of calcium oxide and the mass of magnesium oxide, in grams, obtained from the burning of 1 kg of limestone are, respectively:

a) 60; 21.
b) 100; 84.
c) 184; 96.
d) 336; 100.
e) 600; 210.

Homework Equations

The Attempt at a Solution

Molar masses:
CaCO₃: 40 + 12 + 3*16 = 52 + 48 = 100 g.
MgCO₃: 24,5 + 12 + 3*16 = 36,5 + 48 = 84,5 g.
CaO: 40 + 16 = 56 g.
Mass of CaCO₃ and MgCO₃ in 1 kg of dolomitic limestone (1000 g):
Mass of CaCO₃: 60% of 1000 g = 600 g.
Mass of MgCO₃: 21% of 1000 g = 210 g.
Discover what is the limiting reagent:
[tex]1 mol CaCO_3 \rightarrow 1 mol MgCO_3[/tex]
[tex]100 g \rightarrow 84,5 g[/tex]
[tex]600 g \rightarrow x[/tex]
x = 84,5 * 6 = 507 g of MgCO₃ for consuming all of the CaCO₃; since we only have 210 g:
MgCO₃ => limiting.
Use the mass of MgCO₃ to discover how many CaO will be produced:
[tex]1 mol MgCO_3 \rightarrow 1 mol CaO[/tex]
[tex]84,5 g \rightarrow 56 g[/tex]
[tex]210 g \rightarrow y[/tex]
[tex]y = \frac{210 \times 56}{84,5}[/tex]
y = 139,17 g.
But this mass is not in the answer options. According to the book, the right answer is D (336 g of CaO and 100 g of MgO). What is wrong, then?
Thank you in advance.

My guess is that you need to work with moles when finding the limiting reagent.

 

[pc2-brazil]

Thank you for the response. Actually, we found the solution for this problem in the internet. The solution is completely different from what we imagined:

[tex]1 mol CaCO_3 \rightarrow 1 mol CaO[/tex]
[tex]100 g \rightarrow 56 g[/tex]
[tex]600 g \rightarrow x[/tex]
x = 56 * 6 = 336 g.
[tex]1 mol MgCO_3 \rightarrow 1 mol MgO[/tex]
[tex]84 g \rightarrow 40 g[/tex]
[tex]210 g \rightarrow y[/tex]
y = 210*40/84 = 100 g.
answer: D) 336; 100.

It is strange, because it relates the mass of CaCO₃ with the mass of CaO and the mass of MgCO₃ with the mass of MgO, but doesn't even check what is the limiting reagent.
Any ideas?
Note: we are sure that our check of what is the limiting reagent is right, because we have used the same reasoning in several other problems.

Thank you in advance.

 

[Borek]

This is not a limiting reagent question. CaCO₃ doesn't react with MgCO₃, there are two separate decomposition reactions taking place.

 

[GCT]

Right , I did not read into the question there. It is a bit misleading to place two separate reactions into one stoichiometric equation ; the emphasis here being that such an equation is meant to signify stoichiometric relations despite the fact that there are none here actually.

 

[pc2-brazil]

thank you for the responses. This question is misleading, because the equation gives the idea that there is a reaction between MgCO₃ and CaCO₃.