If 58 grams of Calcium Carbonate is made to react with 40g sulphuric acid, the weight of calcium sulfate obtain is 48 grams. CAlculate the theoretical yield of Calcium Sulfate.

Answer 1

The first step in solving a problem like this is to write out the balanced chemical reaction.

CaCO₃ + H₂SO₄ --> CaSO₄ + H₂CO₃

It is important to make sure we balance the reaction properly, because knowing what the coefficients are before each compound will tell us about the molar ratio.

In other words, we now know that there is a 1:1:1 molar ratio between CaCO₃, H₂SO₄ and CaSO₄ (the product we're interested in). What this means is that 1 mol of each of the reactants is necessary to form 1 mol of product. In chemistry, we're interested in the number of mols of stuff (number of atoms) that react, not necessarily how much those atoms weigh (how many grams that stuff weighs). But we'll talk more about that later.

We need to know what the limiting reactant is in this problem. We can do this by converting both reactants (compounds on the left side of the arrow) from grams into mols. Remember, we're interested in how many mols of atoms react, not necessarily how much those atoms weigh.

We can convert from grams into mols by using the molar mass of both compounds. We can find the molar mass on the periodic table, just underneath the element name.

For CaCO₃, the molar mass is (40.08 g/mol) + (12.01 g/mol) + (15.99 g/mol)*3 (because there are 3 oxygen atoms in this molecule). That equals 100.06 g/mol.

For H₂SO₄, the molar mass is (1.01 g/mol)*2 (because there are 2 hydrogen atoms) + (32.07 g/mol) + 15.99 g/mol)*4 (because there are 4 oxygens. This equals 98.05 g/mol.

The number of mols = mass/molar mass.

So the number of mols of CaCO₃ = 58 g/100.06 g/mol = 0.5797 mols.

And the number of mols of H₂SO₄ = 40 g/98.05 g/mol = 0.4080 mols.

Earlier, when we balanced the chemical reaction, we saw a 1:1 ratio between both reactants and the product. In other words, if I have 0.5797 mols of CaCO₃ and an excess amount of H₂SO₄, then I should be able to make 0.5797 mols of CaSO₄. If I have 0.4080 mols of H₂SO₄ and an excess amount of CaCO₃, then I should be able to make only 0.4080 mols of product. Because I have fewer number of mols of H₂SO₄, I know that there's no way I can make 0.5797 mols of product -- I can only make 0.4080 mols of product at most. The H₂SO₄ is, therefore, our limiting reactant.

In other words, let's say I'm trying to make pancakes (our CaSO₄). The recipe says I need 1 cups of flour and 1 cup of milk (1 mol of CaCO₃ and 1 mol of H₂SO₄) to make 1 batch of pancakes (1 mol of CaSO₄). If I had 3 cups of flour in my pantry, but only 2 cups of milk, then I could only make 2 batches of pancakes. In other words, the milk would be a limiting reactant, because it limits how much of my product I can make.

Theoretically, I can only make 0.4080 mols of CaSO₄. This is my theoretical yield in mols. I can convert that into grams by using the molar mass of the compound.

molar mass = 40.08 g/mol (for the Ca) + 32.07 g/mol (for the S) + 4*15.99 g/mol (for the 4 Os) = 136.11 g/mol

mass = molar mass*number of mols, so:

mass = 0.4080 mols*136.11 g/mol = 55.53 g CaSO₄ produced. The problem tells us that we actually only made 48 g of this product. Theoretically, we could make up to 55.53 g, but in actuality, the process isn't 100% efficient, so we only made 48 g. With the proper number of significant figures, your answer will be 56 g.

Answer 2

Stoichiometry

Write the skeleton equation (formulas): CaCO₃ + H₂SO₄ → CaSO₄ + H₂CO₃

↓

(H₂CO₃ (carbonic acid) decomposes into CO₂ + H₂O) CO₂ + H₂O

The overall skeleton equation is: CaCO₃ + H₂SO₄ → CaSO₄ + CO₂ + H₂O

This equation is also balanced as is, no need for coefficients.

Write the equation, placing the quantities given below the substances in the equation, and the molar masses above the substances in the equation (shown below).

The only relationship between the compounds that is known from the balanced equation is in moles (the coefficients). In this case, 1 mole of calcium carbonate reacts with 1 mol sulfuric acid to produce 1 mol calcium sulfate and 1 mol carbon dioxide and 1 mol water. This means that you have to convert the masses given to moles using the molar mass (shown below).

Molar masses: 100 g/mol 98 g/mol 136 g/mol

CaCaSO₄ + H₂SO₄ —→ CaSO₄ + CO₂ + H₂O

Given: 58 g 40 g ?

Asked:

Because you are given the quantities of 2 substances, you must determine if one is the "limiting reactant". To do this, convert each quantity given from mass to mol, use the ratios from the balanced equation to determine if one of the given quantities makes less product than the other. If so, the one making the least amount of product is the "limiting reactant"

Convert g → mol by dimensional analysis using the molar mass:

58 g CaCO₃ x 1 mol CaCO₃ = 0.58 mol CaCO₃

100 g CaCO₃

Do stoichiometry (calculate amount product) using molar ratio from the balanced equation mol product

mol reactant

0.58 mol CaCO₃ x 1 mol CaSO₄ = 0.58 mol CaSO₄

1 mol CaCO₃

Do same for other quantity of reactant given:

40 g H₂SO₄ x 1 mol H₂SO₄ = 0.41 mol H₂SO₄ x 1 mol CaSO₄ = 0.41 mol CaSO₄

1 mol H₂SO₄

The 40 g H₂SO₄ makes less product, (0.41 mol CaSO₄), than the 58 g CaCO₃ (0.58 CaSO₄) , therefore the H₂SO₄ is the limiting reactant, and the CaCO₃ is present in excess. Any calculations for theoretical yield must be based on the limiting reactant, not the excess reactant.

The THEORETICAL yield is the CALCULATED amount of product that CAN be made, and it is usually reported in g, not mol. The experimental yield, the mass of product ACTUALLY MADE in the lab is compared to the CALCULATED THEORETICAL yield to determine the PERCENT YIELD of the lab reaction.

The last step, then, is to convert the mol CaSO₄ product (calculated from the limiting reactant, H₂SO₄), to g CaSO₄:

0.41 mol CaSO₄ x 136 g CaSO₄ = 56 g CaSO₄ (theoretical yield)

1 mol CaSO₄

To go one more step and calculate the percent yield: 48 g CaSO₄ (actual yield in lab) x 100 = 86%

56 g CaSO₄ (theoretical yield)

Answer 3

CaCO₃ + H₂SO₄ ==> CaSO₄ + CO₂ + H₂O ... balanced equation

Find limiting reactant:

moles CaCO₃ = 58 g x 1 mol/100 g = 0.58 moles

moles H₂SO₄ = 40 g x1 mol/98 g = 0.41 moles

Limiting reactant = H₂SO₄

Theoretical yield CaSO₄ = 0.41 mol H₂SO₄ x 1 mol CaSO₄/mol H₂SO₄ x 136g/mol = 55.8 g = 56 g (to 2 sig figs)