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Usimov [2.4K]
3 years ago
5

Using solution theory and the aide of tables and graphs, explain why the Delta H solution for sodium chloride is favorable to th

e formation of an aqueous solution of sodium chloride.
Chemistry
1 answer:
prisoha [69]3 years ago
4 0
The solution is non-ideal for NaCl soln so ΔH3<span> is greater than the sum of ΔH</span>1<span> and ΔH</span>2. this means the forces of attraction between like molecules is greater than the forces of attraction between unlike molecules. <span>NaCl (table salt) dissolves readily in water. In solid NaCl, the positive sodium ions are attracted to the negative chloride ions. The same is true of the solvent, water; the partially positive hydrogen atoms are attracted to the partially negative oxygen atoms. While NaCl dissolves in water, the positive sodium cations and chloride anions are being stabilized by the water molecule electric dipoles. Thus, the intermolecular bonds between NaCl are broken and the salt is dissolved.</span>
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20.1 g of aluminum and 219 g of chlorine gas react until all of the aluminum metal has been converted to AlCl3. The balanced equ
Romashka-Z-Leto [24]

Answer:

<em>The amount of Cl2 gas left , after the reaction goes to completion is : </em><u><em>139.655 grams</em></u>

Explanation:

Molar mass : It is the mass in grams present in one mole of the substance.

Moles of the substance is calculated by:

Moles=\frac{Mass}{Molar\ mass}

2Al(s)+3Cl_{2}(g)\leftarrow 2AlCl_{3}(g)

According  to this equation:

2 mole of Al = 3 mole of Cl2 = 2 mole of AlCl3

Molar mass of Al = 27.0 g/mol

Mass of Al = 20.1 gram

Moles of Al present in the reaction :

Moles=\frac{Mass}{Molar\ mass}

Moles=\frac{20.1}{26.98}

Moles of Al = 0.744

Similarly calculate the moles of Cl2

Molar mass of Cl2 = 71.0 g/mol

Mass = 219 gram

Moles=\frac{Mass}{Molar\ mass}

Moles=\frac{219}{70.98}

Moles of Cl2 = 3.08 moles

According to equation,

2 mole of Al reacts with = 3 mole of Cl2

1 moles of Al reacts with = 3/2  mole of Cl2

0.744 moles of Al reacts with = 3/2(0.744) moles of Cl2

= 1.116 moles of Cl2

But actually present Cl2 = 3.08 moles

Hence Al is the limiting reagent , and Cl2 is the excess reagent.

The whole Aluminium Al get consumed during the reaction.

The amount of Cl2 in excess = Total Cl2 - Cl2 consumed

Cl2 in excess = 3.08 - 1.116 = 1.964 moles

<u>Cl2 in grams</u><u> </u>= 1.964 x 70.90 <u>= 139.655 grams</u>

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