# Calculate the theoretical yield of carbon dioxide produced from the reaction of 0.500 grams of...

## Question:

Calculate the theoretical yield of carbon dioxide produced from the reaction of 0.500 grams of calcium carbonate and excess hydrochloric acid.

## Acid Rain and Marble Statues:

Many great works of ancient art are made of marble, which is primarily made of calcium carbonate ({eq}\rm CaCO_3 {/eq}). Carbonates are reactive with acids. Increased emissions of sulfur oxides and nitrogen oxides since the industrial revolution contributed to the creation of acid rain, which has increased the rate of damage to marble statues and buildings exposed to rain.

We are asked to calculate the theoretical yield of carbon dioxide when 0.500 g of calcium carbonate ({eq}\rm CaCO_3 {/eq}) reacts with hydrochloric acid (HCl). When an acid reacts with a carbonate, it produces a salt, carbon dioxide ({eq}\rm H_2O {/eq}) and water (H_2O). The salt is an ionic compound made from the metal in the base (in our case, Ca) and the nonmetal in the acid (in our case, Cl). Its chemical formula can be deduced using the criss-cross rule. The balanced neutralization reaction for our reaction is:

{eq}\rm CaCO_3 + 2HCl \to CO_2 + CaCl_2 + H_2O {/eq}

We will now need to convert the mass of calcium carbonate to moles:

{eq}\rm n=\dfrac{m}{M} {/eq}

Where
n = moles
m = mass
M = molar mass

The molar mass of {eq}\rm CaCO_3 {/eq} is obtained by adding the atomic molar masses of each atom. The atomic molar masses are obtained from the periodic table:

{eq}\begin{align} \rm M &= \rm (40.08\:g/mol) + (12.01\:g/mol) + (3 \times 16.00\:g/mol)\\ &= \rm 100.09\:g/mol \end{align} {/eq}

Now we can substitute the values in and solve:

{eq}\begin{align} \rm n &= \rm \dfrac{m}{M}\\ &= \rm \dfrac{0.500\:g}{100.09\:g/mol}\\ &= \rm 0.00500\:mol \end{align} {/eq}

Next we will calculate how many moles of carbon dioxide by multiplying the moles of calcium carbonate by its mole ratio (based on the coefficients from our balanced chemical equation) with carbon dioxide.

{eq}\rm 0.00500\:mol_{CaCO_3} \times \dfrac{1\:mol_{CO_2}}{1\:mol_{CaCO_3}} = 0.00500\:mol_{CO_2} {/eq}

We can now convert the theoretical yield (in moles) of carbon dioxide to mass:

{eq}\begin{align} \rm M_{CO_2} &= \rm (12.01\:g/mol) + (2 \times 16.00\:g/mol)\\ &= \rm 44.01\:g/mol \end{align} {/eq}

Now we can substitute the values in and solve:

{eq}\begin{align} \rm m &= \rm nM\\ &= \rm (0.00500\:mol)(44.01\:g/mol)\\ &= \rm 0.220\:g \end{align} {/eq}

The theoretical yield of carbon dioxide is 0.220 g. 