Chemistry · General chemistry I · Worked example

Theoretical yield of CO₂ from CaCO₃ and HCl

A $5.00 g$ piece of pure ${CaCO}_{3}$ is dropped into acid holding $0.200 mol HCl$ . What mass of ${CO}_{2}$ can form, at most?

Write down the starting amounts and the question

The data are $5.00 g$ of pure ${CaCO}_{3}$ and $0.200 mol$ of $HCl$ , and the question is the theoretical mass of ${CO}_{2}$ . Assume the stated reaction goes to completion and that no other reaction uses these reactants. The $HCl$ is already in moles, so 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 ${CaCl}_{2}$ and the two hydrogens in $H_{2} O$ . Keep the subscripts unchanged. The state symbols mean solid (s), aqueous (aq), liquid (l) and gas (g).

{CaCO}_{3} (s) + 2 HCl (aq) \to {CaCl}_{2} (aq) + H_{2} O (l) + {CO}_{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. The coefficients can now be read as ratios. $1 mol {CaCO}_{3}$ needs $2 mol HCl$ and produces $1 mol {CO}_{2}$ , and these are the mole ratios the calculation uses.

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

Calculate the two molar masses

Add the atomic mass of every atom in each formula. Keep the full values through the work and round only 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 decimal shown is rounded for reading, so 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 ${CaCO}_{3}$ amount by the $2$ : $1$ ratio of $HCl$ to ${CaCO}_{3}$ . The carbonate needs about $0.0999101 mol$ of $HCl$ , less than the $0.200 mol$ available. $HCl$ is in excess, so ${CaCO}_{3}$ 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. ${CaCO}_{3}$ allows about $0.0499550 mol$ of reaction, and $HCl$ allows $0.200 / 2 = 0.100 mol$ . The smaller amount is the limit, which confirms that the carbonate runs out first.

Connect carbonate to carbon dioxide

Now use the $1$ : $1$ mole ratio. The available carbonate gives about $0.0499550 mol {CO}_{2}$ . Equal amounts in moles do not mean equal masses, because ${CaCO}_{3}$ and ${CO}_{2}$ 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 $2.20 g {CO}_{2}$ 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 ${CO}_{2}$ , and the rest goes into the other products. Your own calculator, carrying every digit, should give the same $2.20 g$ .

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

Result

The theoretical ${CO}_{2}$ yield is $2.20 g$ . ${CaCO}_{3}$ is limiting.

Your turn

1. One like the example

With $HCl$ in excess, what is the theoretical mass of ${CO}_{2}$ from $12.5 g$ of pure ${CaCO}_{3}$ ?

Show the answer and explanation

$5.50 g$ of ${CO}_{2}$ .

The acid is in excess, so the carbonate limits. $12.5 g \div 100.09 g / mol$ $= 0.1249 mol$ of ${CaCO}_{3}$ , which makes $0.1249 mol$ of ${CO}_{2}$ by the $1$ : $1$ ratio. $0.1249 \times 44.01 = 5.50 g$ .

\begin{array}{l} {CaCO}_{3} (s) + 2 HCl (aq) \to {CaCl}_{2} (aq) + H_{2} O (l) + {CO}_{2} (g) \\ 12.5 g {CaCO}_{3} \times \frac{1 mol {CaCO}_{3}}{100.09 g {CaCO}_{3}} \times \frac{1 mol {CO}_{2}}{1 mol {CaCO}_{3}} \times \frac{44.01 g {CO}_{2}}{1 mol {CO}_{2}} = 5.50 g {CO}_{2} \end{array}

2. A variation

Keep $HCl$ at $0.200 mol$ , but increase ${CaCO}_{3}$ to $15.0 g$ . Which reactant limits the reaction, and what is the theoretical mass of ${CO}_{2}$ ?

Show the answer and explanation

$HCl$ is limiting, and the theoretical ${CO}_{2}$ yield is $4.40 g$ .

$15.0 g$ of carbonate would need $0.300 mol$ of $HCl$ , but only $0.200 mol$ is there, so the acid runs out first. $0.200 mol HCl$ makes $0.100 mol {CO}_{2}$ by the $1$ : $2$ ratio, which is $4.401 g$ , or $4.40 g$ to three figures.

\begin{array}{l} {CaCO}_{3} (s) + 2 HCl (aq) \to {CaCl}_{2} (aq) + H_{2} O (l) + {CO}_{2} (g) \\ 15.0 g {CaCO}_{3} \times \frac{1 mol {CaCO}_{3}}{100.09 g {CaCO}_{3}} \times \frac{2 mol HCl}{1 mol {CaCO}_{3}} = 0.300 mol HCl \\ 0.200 mol HCl \times \frac{1 mol {CO}_{2}}{2 mol HCl} \times \frac{44.01 g {CO}_{2}}{1 mol {CO}_{2}} = 4.40 g {CO}_{2} \end{array}

3. Somewhere new

What mass of ${CaCO}_{3}$ reacts completely with $25.0 mL$ of $2.00 M HCl$ ?

Show the answer and explanation

$2.50 g$ of ${CaCO}_{3}$ .

The acid is given as a solution, so its moles come from $n = M V$ $= (2.00 M) (0.0250 L)$ $= 0.0500 mol$ . Two $HCl$ react with each ${CaCO}_{3}$ , so $0.0250 mol$ of carbonate reacts, and $0.0250 \times 100.09 = 2.50 g$ .

\begin{array}{l} {CaCO}_{3} (s) + 2 HCl (aq) \to {CaCl}_{2} (aq) + H_{2} O (l) + {CO}_{2} (g) \\ n = M V \\ M = 2.00 M \\ V = 0.0250 L \\ n = (2.00 M) (0.0250 L) = 0.0500 mol \\ 0.0500 mol HCl \times \frac{1 mol {CaCO}_{3}}{2 mol HCl} \times \frac{100.09 g {CaCO}_{3}}{1 mol {CaCO}_{3}} = 2.50 g {CaCO}_{3} \end{array}

Keep exploring

Open the mixture in Stoichiometry & yield and raise the ${CaCO}_{3}$ to $15.0 g$ : the acid now limits, and the yield stops at $4.40 g$ of ${CO}_{2}$ .

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After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 4 key terms Explore Stoichiometry & yield Graph yield with excess acid Check the chain in a Chemistry box Open worked example on a board Periodic table and atomic masses

Your existing work stays on this device. Examples open as editable copies.

Write down the starting amounts and the question

Write and balance the reaction

Check the atoms and read the ratios

Calculate the two molar masses

Convert the carbonate mass to moles

Check whether there is enough acid

Connect carbonate to carbon dioxide

Convert the product to grams

Report and interpret the result

Result

Your turn

1. One like the example

5.50 g5.50 g of CO2CO2​.

2. A variation

HClHCl is limiting, and the theoretical CO2CO2​ yield is 4.40 g4.40 g.

3. Somewhere new

2.50 g2.50 g of CaCO3CaCO3​.

Keep exploring

Study it, then try it

$5.50 g$ of ${CO}_{2}$ .

$HCl$ is limiting, and the theoretical ${CO}_{2}$ yield is $4.40 g$ .

$2.50 g$ of ${CaCO}_{3}$ .