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Ronch [10]
3 years ago
7

Jamal is working with three ionic compounds: sodium chloride, calcium sulfide, and barium oxide. His teacher asks him which are

the positive ions in each compound. What should Jamal’s answer be? sodium, calcium, and barium barium, sulfur, and oxygen calcium, barium, and chlorine chlorine, sulfur, and oxygen
Chemistry
1 answer:
Lubov Fominskaja [6]3 years ago
4 0

Answer:

option A is correct  ( sodium, calcium and barium)

Explanation:

Given compounds:

Sodium chloride , calcium sulfide, barium oxide

We know that metals form positive ions. In order to solve the problem we must identify the metals from given compounds.

Na⁺Cl⁻

Ca²⁺S²⁻

Ba²⁺O²⁻

We can see that sodium, calcium and barium contain positive charges.

Thus option A is correct.

Because sodium have one valance electron. When it combine with chlorine sodium lose its one electron to complete the octet and chlorine accept it to complete its octet. Thus sodium form positive ion and chlorine form negative ion.

Similarly barium and calcium are present in group 2. Both have two valance electron. When they lose them cation are formed.

Other option are incorrect because,

Option B have sulfur and oxygen which are anion.

Option C have chlorine which is also anion

Option D have chlorine, sulfur and oxygen that are anions.

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In a reaction, chlorine accepts an electron to form Cl-. What is true about the atomic size of the chlorine ion?
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Answer is: The atomic size of the chlorine ion is larger than the size of the chlorine atom.

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Limiting Reactants—————-
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Answer:

21.8 grams.

Explanation:

Molar mass data from a modern periodic table:

  • Mg: 24.301;
  • O: 15.999.

How many moles of MgO will be produced if Mg is the limiting reactant?

Number of moles of Mg:

\displaystyle n = \frac{m}{M} = \frac{16.3}{24.301} = 0.670644\;\text{mol}.

The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.

How many moles of MgO will be produced if O₂ is the limiting reactant?

Number of moles of O₂:

\displaystyle n = \frac{m}{M} = \frac{4.33}{15.999} = 0.270642\;\text{mol}.

The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. 2\times 0.270642 = 0.541284\;\text{mol} of MgO will be produced if O₂ is in excess.

How many moles of MgO will be produced?

0.541284 is smaller than 0.670644. Only 0.541284 moles of MgO will be produced since O₂ will run out before all 16.3 grams of Mg is consumed.

What's the mass of 0.541284 moles of MgO?

Formula mass of MgO:

24.301 + 15.999 = 40.300\;\text{g}\cdot\text{mol}^{-1}.

Mass of 0.541284 moles of MgO:

m = n \cdot M = 0.541284\times 40.300 = 21.8\;\text{g}.

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