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pychu [463]
3 years ago
8

Calcium reacts with bromine to form calcium bromide. 2.1 Which atom will lose electrons? 2.2 How many electrons will it lose? 2.

3 Name the element that will gain electrons. 2.4 How many electrons does this atom gain? 2.5 Write down the symbol for each of the ions formed. 2.6 Write the chemical formula for calcium bromide.
Chemistry
1 answer:
Juli2301 [7.4K]3 years ago
8 0

<u>Answer:</u>

<u>For 2.1:</u> Calcium atom will loose electrons.

<u>For 2.2:</u> It will loose 2 electrons.

<u>For 2.3:</u> Bromine atom will gain electrons.

<u>For 2.4:</u> It will gain 1 electron.

<u>For 2.5:</u> The ions formed are Ca^{2+}\text{ and }Br^-

<u>For 2.6:</u> The chemical formula is CaBr_2

<u>Explanation:</u>

Calcium bromide is formed by the combination of calcium and bromine atoms. This compound is ionic compound because electrons get transferred from one atom to another atom.

Calcium is the 20th element of the periodic table. It is a metal and it looses its electrons. The electronic configuration for this element is 1s^22s^22p^63s^23p^64s^2

This element will loose 2 electrons and forms Ca^{2+} ion.

Bromine is the 35th element of the periodic table. It is a non-metal and it gains electrons. The electronic configuration for this element is [Ar]4s^23d^{10}4p^5

This element will gain 1 electron and forms Br^- ion.

The chemical formula for the given compound is CaBr_2

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Determine the LIMITING reactant in the following balanced equation:
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Answer:

KBr is limiting reactant.

Explanation:

Given data:

Mass of  KBr =4g

Mass of Cl₂ = 6 g

Limiting reactant = ?

Solution:

Chemical equation:

2KBr + Cl₂      →    2KCl + Br₂

Number of moles of KBr:

Number of moles = mass/molar mass

Number of moles = 4 g/ 119 gmol

Number of moles = 0.03 mol

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 6 g/ 70 gmol

Number of moles = 0.09 mol

Now we will compare the moles of reactant with product.

              KBr            :            KCl

                2              :              2

            0.03            :            0.03

             KBr            :              Br₂

                2             :               1

             0.03           :          1/2×0.03= 0.015

               Cl₂             :            KCl

                 1              :              2

            0.09            :           2/1×0.09 = 0.18

               Cl₂             :              Br₂

                1              :               1

             0.09           :            0.09

Less number of moles of product are formed by the KBr thus it will act as limiting reactant while Cl₂  is present in excess.

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