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28 August, 03:57

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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  1. 28 August, 04:20
    0
    Mass of CaCl₂ produced = 15 g

    Excess reactant = CaCO₃

    Mass of CaCO₃ left = 11.5 g

    Explanation:

    Given dа ta:

    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 CaCl₂ 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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