Question

Using the equationCaCO3 (s) + 2 HCl (aq) arrow CaCl2 + CO2 (g) + H2O (l)calculate the molarity of a hydrochloric acid solution if an initial mass of 5.238 g of CaCO3 was reacted with 25.0 mL of the of the acid and 3.77 g of CACO3 remain after the reaction is complete.

Answer 1

Answer:

1.17 M HCl.

Explanation:

What is given?

Mass of CaCO3 = 5.238 g.

Excess of CaCO3 = 3.77 g.

Mass of CaCO3 consumed in the reaction = 5.238 g - 3.77g = 1.468 g.

Molar mass of CaCO3 = 100 g/mol.

Volume of HCl solution = 25.0 mL.

Step-by-step solution:

As you can see, we have an excess of CaCO3 because we have 3.77g of CaCO3 remaining and initially we used 5.238 g of this compound that actually was consumed 1.468 g in the reaction. So, let's see how many moles there are in 1.468 g of CaCO3 using its molar mass:

$1.468\text{ g CaCO}_3\cdot\frac{1\text{ mol CaCO}_3}{100\text{ g CaCO}_3}=0.01468\text{ moles CaCO}_3.$

And now, based on this, you can see that in the chemical equation, we have 1 mol of CaCO3 reacting with 2 moles of HCl, so let's find how many moles of HCl we need to react with 0.01468 moles of CaCO3:

$0.01468\text{ moles CaCO}_3\cdot\frac{2\text{ moles HCl}}{1\text{ mol CaCO}_3}=0.02936\text{ moles HCl.}$

Now that we have the number of moles of HCl and its volume (25.0 mL), we can calculate its molarity using the following formula:

$Molarity=\frac{mole\text{s of solute}}{liter\text{s of solution}}=\frac{mol}{L}.$

But we need to have the volume in liters. Remember that 1 L equals 1000 mL, so:

$25.0\text{ mL}\cdot\frac{1\text{ L}}{1000\text{ mL}}=0.025\text{ L.}$

And finally, we can replace our data with the molarity formula:

$Molarity\text{ of HCl=}\frac{0.02936\text{ moles}}{0.025\text{ L}}=1.174\text{ M.}$

The molarity of the hydrochloric acid solution would be 1.17 M.