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When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.
CaCO3 (s) + 2HCL (aq) → CaCl2 (aq) + H2O (l) + CO2 (g)

How many grams of calcium chloride will be produced when 25.0 g of calcium carbonate is combined with 10.0 g of hydrochloric acid?

Which reactant is in excess? HCl or CaCO3?

How many grams of the excess reactant will remain after the reaction is complete?

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Answer:

Mass of CaClâ‚‚ produced = 15 g

Excess reactant = CaCO₃

Mass of CaCO₃ left = 11.5 g

Explanation:

Given data:

Mass of calcium carbonate = 25 g

Mass of hydrochloric acid = 10.0 g

Mass of calcium chloride produced = ?

Chemical equation:

CaCO₃ + 2HCl  → CaCl₂  + H₂O + CO₂

Number of moles of CaCO₃:

Number of moles of CaCO₃ = Mass /molar mass

Number of moles of CaCO₃= 25.0 g / 100.1 g/mol

Number of moles of CaCO₃ = 0.25 mol

Number of moles of HCl:  

Number of moles of HCl = Mass /molar mass

Number of moles of HCl = 10.0 g / 36.5 g/mol

Number of moles of HCl = 0.27 mol

Now we will compare the moles of CaCl₂ with HCl and CaCO₃ .

                  CaCO₃      :               CaCl₂

                   1               :               1

                 0.25           :            0.25

                HCl              :                CaCl₂

                 2                :                 1

                 0.27            :               1/2 × 0.27 = 0.135 mol

The number of moles of CaClâ‚‚ produced by HCl are less it will be limiting reactant.

Mass of CaCl₂ = moles × molar mass

Mass of CaCl₂ =0.135 mol × 110.98 g/mol

Mass of CaCl₂ =  15 g

The calcium carbonate is present in excess.

               HCl              :                CaCO₃

                 2                :                 1

                 0.27            :               1/2 × 0.27 = 0.135 mol

So, 0.135 moles react with 0.27 moles of HCl.

The moles of CaCO₃ remain unreacted = 0.25 - 0.135

The moles of CaCO₃ remain unreacted = 0.115 mol

Mass remain unreacted:

Mass of of CaCO₃ remain unreacted = Moles × molar mass

Mass of of CaCO₃ remain unreacted = 0.115 mol × 100.1 g/mol

Mass of of CaCO₃ remain unreacted = 11.5 g

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