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Doss [256]
3 years ago
5

Classify each of these soluble solutes as a strong electrolyte, a weak electrolyte, or a nonelectrolyte. Solutes formula hydroch

loric acid hcl sodium hydroxide naoh formic acid hcooh methyl amine ch3nh2 potassium chloride kcl ethanol c2h5oh sucrose c12h22o11
Chemistry
1 answer:
matrenka [14]3 years ago
3 0

Compounds which on dissolving in water gets completely dissociates into its ions are known as strong electrolytes whereas compounds which on dissolving in water gets partially dissociates into its ions are known as weak electrolytes.


Substances which gives solution on dissolving in water and do not dissociates into ions also does not conduct electric current are known as nonelectrolyte.

  • Hydrochloric acid, HCl

On adding HCl (strong acid) in water, it will completely dissociates into ions (H^{+} and Cl^{-}) and thus, it is a strong electrolyte.

  • Sodium hydroxide, NaOH

On adding NaOH (strong base) in water, it will completely dissociates into ions (Na^{+} and  OH^{-}) and thus, it is a strong electrolyte.

  • Formic acid, HCOOH

On adding HCOOH (weak acid) in water, it will partially dissociates into ions (H^{+} and  HCOO^{-}) and thus, it is a weak electrolyte.

  • Methyl amine, CH_3NH_2

On adding CH_3NH_2 (weak base) in water, it will partially dissociates into ions (CH_3NH_3^{+} and  OH^{-}) and thus, it is a weak electrolyte.

  • Potassium chloride, KCl

On adding KCl in water, it will completely dissociates into ions (K^{+} and  Cl^{-}) and thus, it is a strong electrolyte.

  • Ethanol, C_2H_5OH

On adding C_2H_5OH in water, it will not dissociates into ions  and thus, it is a nonelectrolyte.

  • Sucrose, C_{12}H_{22}O_{11}

On adding C_{12}H_{22}O_{11} in water, it will not dissociates into ions  and thus, it is a nonelectrolyte.

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The percent yield of the calcium hydroxide is 84.5%.

<h3>What is stoichiometry?</h3>

Stoichiometry enables us to obtain the mass of a substance form the equation of the reaction.

The equation of the reaction is;

CaCO3 + 2HCl -----> CaCl2 + CO2 + H2O

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The reaction is 1:1 then the amount of CO2  produced is 0.4 moles

Mass of CO2 = 0.4 CO2 * 44 g/mol = 17.6 g

2) The reaction equation is; 2NaOH + CaCO3 --->  Ca(OH)2 + Na2CO3

Number of moles of X = 25.0 grams/100 g/mol =  0.25 moles

Number of moles of NaOH= 40/1000 L * 2 M = 0.08 moles

If 1 mole of X reacts with 2 moles of NaOH

0.25 moles  reacts with   0.25 moles   * 2 moles /1 mole

= 0.5 moles

NaOH is the limiting reactant

2 moles of NaOH produces 1 mole of CO2

0.08 moles of NaOH produces 0.08 moles * 1 mole/2 moles

= 0.04 moles of CO2

Theoretical yield of CO2 =  0.04 moles of CO2 * 74 g/mol = 2.96  g

Percent yield = 2.5 g/ 2.96  g * 100

= 84.5%

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