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avanturin [10]
4 years ago
8

Dacă intr-o scoica se introduc câteva picături de acid se observa apariția efervescenței. Scoica conține carbonat de calciu.Daca

s-au introdus 1,46g de acid clorhidric ce cantități de produși de reacție se obțin?
Chemistry
1 answer:
Step2247 [10]4 years ago
5 0

Answer:

Explanation:

In English;

If a few drops of acid are introduced into a shell, the effervescence may be observed. The shell contains calcium carbonate.If 1.46g of hydrochloric acid have been introduced, what quantities of reaction products are obtained?

Given parameters:

Mass of HCl = 1.46g

Unknown:

Mass of CO₂ produced = ?

Mass of H₂O produced =?

Mass of CaCl₂ produced = ?

Solution

 The balanced reaction equation is given below:

             CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

The effervescence observed is due to the production of carbon dioxide gas.

In order to find the mass of each of the products obtain, we must first express the known mass of the acid in terms of moles:

 number of moles of HCl = \frac{mass of HCl}{molar mass of HCl}

  Molar mass of HCl = 1 + 35.5 = 36.5g/mol

  number of moles of HCl = \frac{1.46}{36.5} = 0.04mole

Now, we must find the number of moles of each of the products obtained

number of moles of CaCl₂:

      2 moles of HCl produced 1 mole of CaCl₂

       0.04 moles of HCl will produce  \frac{0.04}{2} moles = 0.02moles of CaCl₂

Also, number of mole of H₂O is 0.02moles

        number of mole of CO₂ is 0.02moles

Mass of the products:

  Mass of CaCl₂ = number of moles of CaCl₂ x molar mass of CaCl₂

        molar mass of CaCl₂ = 40 + 2(35.5) = 111g/mol

         mass of CaCl₂ = 0.02 x 111 = 2.22g

Mass of H₂O = number of moles of H₂O x molar mass of H₂O

        molar mass of H₂O = 2(1) + 16= 18g/mol

        mass of H₂O = 0.02 x 18 = 0.36g

Mass of CO₂ = number of moles of CO₂ x molar mass of CO₂

        molar mass of CO₂ = 12 + 2(16) = 44g/mol

         mass of CO₂ = 0.02 x 44 = 0.88g

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What is Half Life ?

Half-life is the time required by an unstable element  to reduce to half of its initial value.

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The formula is

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