Answer :
Taking into account the reaction stoichiometry and limiting reagent, 26.72 grams of Naâ‚‚SOâ‚„ are formed when 22 grams of NaOH reacts with 21 grams of Hâ‚‚SOâ‚„.
Reaction stoichiometry
In first place, the balanced reaction is:
2 NaCl + H₂SO₄ → Na₂SO₄ + 2 HCI
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- NaCl: 2 moles
- Hâ‚‚SOâ‚„: 1 mole
- Naâ‚‚SOâ‚„: 1 mole
- HCI: 2 moles
The molar mass of the compounds is:
- NaCl: 58.45 g/mole
- Hâ‚‚SOâ‚„: 98 g/mole
- Naâ‚‚SOâ‚„: 142 g/mole
- HCI: 36.45 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
NaCl: 2 moles ×58.45 g/mole= 116.9 grams
H₂SO₄: 1 mole ×98 g/mole= 98 grams
Na₂SO₄: 1 mole ×142 g/mole= 142 grams
HCI: 2 moles ×36.45 g/mole= 72.9 grams
Limiting reagent
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
Limiting reagent in this reaction
To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 116.2 grams of NaCl reacts with 98 grams of Hâ‚‚SOâ‚„, 22 grams of NaCl reacts with how much moles of Hâ‚‚SOâ‚„?
[tex]mass of H_{2} SO_{4} =\frac{22 grams of NaClx98 grams of H_{2} SO_{4} }{116.2 grams of NaCl}[/tex]
mass of Hâ‚‚SOâ‚„= 18.55 grams
But 18.55 grams of Hâ‚‚SOâ‚„ are not available, 21 grams are available. Since you have less moles than you need to react with 22 grams of NaCl, Hâ‚‚SOâ‚„ will be the limiting reagent.
Mass of Naâ‚‚SOâ‚„ produced
The following rules of three can be applied, considering the limiting reagent: if by reaction stoichiometry 116.9 grams of NaCl form 142 grams of Naâ‚‚SOâ‚„, 22 grams of NaCl form how much mass of Naâ‚‚SOâ‚„?
[tex]mass of Na_{2}S O_{4} =\frac{22 grams of NaClx142 grams of Na_{2}S O_{4}}{116.9 grams of NaCl}[/tex]
mass of Naâ‚‚SOâ‚„= 26.72 grams
Then, 26.72 grams of Naâ‚‚SOâ‚„ are formed when 22 grams of NaOH reacts with 21 grams of Hâ‚‚SOâ‚„. The correct answer is first option.
Learn more about the reaction stoichiometry:
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