Question

A tablet contains CaCO3 and reacted with excess hydrochloric acid to produce 0.44g of Carbon Dioxide. CaCO3 + 2HCl = CaCl2 + CO2 + H2O MgCO3 + 2HCl = MgCl2 + CO2 + H2O How many moles of CaCO3 are in a tablet?

Please explain how knowing the mole of CO2 can help in knowing the moles of CaCO3. How does the equation show that the mole of co2 should be halved to result in the moles of caco3?

Answer 1

Look at the stoichiometry of your equation.....

Explanation:

`CaCO_3(aq)+2HCl(aq) rarr CaCl_2(aq) + H_2O(l) +CO_2(g)uarr`

Now this equation is stoichiometrically balanced, and it tells us that one mole of calcium carbonate, an `100*g` mass, reacts with excess acid to give ONE MOLE of calcium chloride, and ONE MOLE of carbon dioxide....the mass of the products and the mass of the reactants are NECESSARILY EQUAL..because masses are conserved in all chemical reaction.

So given the stoichiometric equation, the molar equivalence, if I know the amount, the moles of one reactant or product, then I know the mass, the moles of the other reactants and products. Are you with me? Typically one reagent will be in excess, and the reagent in deficiency is known as the limiting reagent, and the stoichiometry depends on the molar quantity of this limiting reagent.

Should we burn hydrocarbons in air, for instance, hexanes...`C_6H_14`, then the hydrocarbon is the limiting reagent...and we work out molar equivalence based on this reagent:

`C_6H_14(l) + 19/2O_2(g) rarr 6CO_2(g) + 7H_2O(l) + Delta`

What does the `Delta` symbol mean here?

Stoichiometry is just a fancy word for saying `"garbage in equal garbage out"`; it literally means `"in the right measure"`. We practise stoichiometry all the time, for instance when we pay for something with a large banknote, and the seller makes change. Are you often short-changed? I don't think so, because we are well-trained to keep track of our money. As a chemist, you train to keep track of your reactants and products.

And here we gots `(0.44*g)/(44.01*g*mol^-1)=0.010*mol` with respect to carbon dioxide. Necessarily, given the stoichiometry, AT MOST there was a molar quantity of `0.010*mol` with respect to calcium carbonate....

And a mass of `0.01*molxx100.08*g*mol^-1=1.00*g` with respect to the carbonate.

See here for more of the same.

Please review your question and this answer, and if there are queries or objections, voice them, and someone will help you.

Answer 2

Refer to the explanation.

Explanation:

Balanced equation

`"CaCO"_3 + "2HCl"` `rarr` `"CaCl"_2 + "CO"_2 + "H"_2"O"`

Start with carbon dioxide and work backwards to calcium carbonate.

The molar mass of `"CO"_2` is needed.

Multiply the subscript of each element in the compound by its molar mass (atomic weight on the periodic table in g/mol).

`"CO"_2":` `(1xx"12.011 g/mol C") + (2xx"15.999 g/mol O") = "44.009 g/mol CO"_2`

Convert the given mass of `"CO"_2` to moles by dividing by its molar mass. Since molar mass is a fraction `("g"/"mol")`, you can divide by multiplying by its reciprocal `("mol"/"g")`.

Next multiply mol `"CO"_2` by the mole ratio between `"CaCO"_3` and `"CO"_2` from the balanced equation, so that `"CaCO"_3` will be in the numerator. This will give mol `"CaCO"_3` in the tablet.

This is an important step even if the mol ratios are the same, because this step is where we go from mol `"CO"_2` to mol `"CaCO"_3`.

The equation does not indicate mol `"CO"_2` will be halved in order to determine mol `"CaCO"_3` because the mol ratio between them is `1:1`. The mol `"CO"_2` and `"CaCO"_3` are both `"0.010 mol"` rounded to two significant figures.

`0.44color(red)cancel(color(black)("g CO"_2))xx(1color(red)cancel(color(black)("mol CO"_2)))/(44.009color(red)cancel(color(black)("g CO"_2)))xx(1"mol CaCO"_3)/(1color(red)cancel(color(black)("mol CO"_2)))="0.010 mol CaCO"_3` rounded to two significant figures