Chemistry · General chemistry · Worked example

Predict CO₂ from calcium carbonate

Model 5.00 g of pure CaCO₃ with 0.200 mol HCl. What is the theoretical mass of CO₂?

Write down the starting amounts and the question

We have 5.00 g of pure CaCO₃ and 0.200 mol HCl. We want the theoretical mass of CO₂. Assume the stated reaction goes to completion and no other reaction consumes these reactants. The HCl amount is already in moles; no solution volume is needed.

m ({CaCO}_{3}) = 5.00 g, n (HCl) = 0.200 mol

Write and balance the reaction

Calcium carbonate and hydrochloric acid form calcium chloride, water and carbon dioxide. Put a coefficient of 2 before HCl to supply the two chlorines in CaCl₂ and the two hydrogens in H₂O. Keep the subscripts unchanged. The state symbols mean solid (s), aqueous (aq), liquid (l) and gas (g).

C a C O_{3} (s) + 2 H C l (a q) \to C a C l_{2} (a q) + H_{2} O (l) + C O_{2} (g)

Check atoms in the balanced reaction
Element	Reactant side	Product side
Ca	1	1
C	1	1
O	3	3
H	2	2
Cl	2	2

Check the atoms and read the ratios

The atom-count table shows the same count of each element on both sides. We can now use the coefficients: 1 mol CaCO₃ needs 2 mol HCl and produces 1 mol CO₂. This is where the mole ratios in the calculation come from.

1 mol {CaCO}_{3} : 2 mol HCl : 1 mol {CO}_{2}

Calculate the two molar masses

Using the bundled Tro atomic masses, add the contribution from each atom. We will use the full calculated values during the work and round the final yield.

\begin{aligned} M ({CaCO}_{3}) & = 40.08 + 12.01 + 3 (16.00) \\ = 100.09 g/mol \\ M ({CO}_{2}) & = 12.01 + 2 (16.00) \\ = 44.01 g/mol \end{aligned}

Convert the carbonate mass to moles

Divide the available mass by its molar mass. Written as a factor, grams cancel and leave moles. The displayed decimal is rounded for reading; keep the full quotient for the final multiplication.

5.00 g {CaCO}_{3} \times \frac{1 mol {CaCO}_{3}}{100.09 g {CaCO}_{3}} \approx 0.0499550 mol {CaCO}_{3}

Check whether there is enough acid

Multiply the CaCO₃ amount by the 2:1 HCl-to-CaCO₃ ratio. We need about 0.0999101 mol HCl, less than the available 0.200 mol. HCl is in excess, so CaCO₃ limits the yield.

0.0499550 mol {CaCO}_{3} \times \frac{2 mol HCl}{1 mol {CaCO}_{3}} \approx 0.0999101 mol HCl

Check by comparing reaction extents

Divide each available amount by its coefficient. CaCO₃ permits about 0.0499550 mol of reaction; HCl permits 0.200/2 = 0.100 mol of reaction. The smaller amount is the limit, confirming our conclusion.

Connect carbonate to carbon dioxide

Now use the 1:1 mole ratio. The available carbonate gives about 0.0499550 mol CO₂. Equal mole amounts do not mean equal masses: CaCO₃ and CO₂ have different molar masses.

n ({CO}_{2}) = \frac{5.00}{100.09} mol \approx 0.0499550 mol

Convert the product to grams

Multiply by 44.01 g/mol. The complete factor chain makes the path from grams of carbonate to grams of carbon dioxide visible. On the board, tap matching units to show which ones cancel.

\frac{5.00}{100.09} mol {CO}_{2} \times \frac{44.01 g {CO}_{2}}{1 mol {CO}_{2}} \approx 2.19852133 g {CO}_{2}

Report and interpret the result

Report 2.20 g CO₂ to three significant figures. That is the theoretical yield, not a measured collection. It is smaller than 5.00 g because only part of the carbonate’s mass ends up in CO₂; the rest goes into other products. The calculator result should agree with the unrounded calculation.

m ({CO}_{2}) = 2.20 g (three significant figures)

Result

The theoretical CO₂ yield is 2.20 g. CaCO₃ is limiting.

Your turn

Keep HCl at 0.200 mol, but increase CaCO₃ to 15.0 g. Which reactant limits the reaction, and what is the theoretical mass of CO₂?

Show the answer and explanation

HCl is limiting, and the theoretical CO₂ yield is 4.40 g.

The carbonate provides about 0.1499 mol, but the acid can react with only 0.200/2 = 0.100 mol CaCO₃. That produces 0.100 mol CO₂, or 4.401 g before rounding. Report 4.40 g to three significant figures. A straight-line graph that assumes excess acid no longer describes this finite-acid case.

0.200 mol HCl \times \frac{1 mol {CO}_{2}}{2 mol HCl} \times \frac{44.01 g {CO}_{2}}{1 mol {CO}_{2}} = 4.401 g {CO}_{2}

Keep exploring

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Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.