# Strange result in stoichiometry limiting/excess problem.

1. Nov 29, 2008

### pc2-brazil

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

1. The problem statement, all variables and given/known data

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:

$$CaCO_3_{(s)} + MgCO_3_{(s)} \rightarrow CaO_{(s)} + MgO_{(s)} + 2CO_2_{(g)}$$

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.

2. Relevant equations

3. 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:
$$1 mol CaCO_3 \rightarrow 1 mol MgCO_3$$
$$100 g \rightarrow 84,5 g$$
$$600 g \rightarrow x$$
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:
$$1 mol MgCO_3 \rightarrow 1 mol CaO$$
$$84,5 g \rightarrow 56 g$$
$$210 g \rightarrow y$$
$$y = \frac{210 \times 56}{84,5}$$
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?

Last edited: Nov 29, 2008
2. Nov 29, 2008

### GCT

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

3. Nov 29, 2008

### 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:

$$1 mol CaCO_3 \rightarrow 1 mol CaO$$
$$100 g \rightarrow 56 g$$
$$600 g \rightarrow x$$
x = 56 * 6 = 336 g.
$$1 mol MgCO_3 \rightarrow 1 mol MgO$$
$$84 g \rightarrow 40 g$$
$$210 g \rightarrow y$$
y = 210*40/84 = 100 g.

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.

4. Nov 29, 2008

### Staff: Mentor

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

5. Nov 29, 2008

### 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.

6. Nov 30, 2008

### 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.