When 22.0 g NaCl and 21.0 g H, SO4 are mixed and react according to the equation below.
1 answer:
Taking into account the reaction stoichiometry and limiting reagent, 26.72 grams of Na₂SO₄ are formed when 22 grams of NaOH reacts with 21 grams of H₂SO₄.
<h3>Reaction stoichiometry</h3>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
- H₂SO₄: 1 mole
- Na₂SO₄: 1 mole
- HCI: 2 moles
The molar mass of the compounds is:
- NaCl: 58.45 g/mole
- H₂SO₄: 98 g/mole
- Na₂SO₄: 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
<h3>Limiting reagent</h3>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.
<h3>Limiting reagent in this reaction</h3>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 H₂SO₄, 22 grams of NaCl reacts with how much moles of H₂SO₄?
<u><em>mass of H₂SO₄= 18.55 grams</em></u>
But 18.55 grams of H₂SO₄ are not available, 21 grams are available. Since you have less moles than you need to react with 22 grams of NaCl, H₂SO₄ will be the limiting reagent.
<h3>Mass of Na₂SO₄ produced</h3>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 Na₂SO₄, 22 grams of NaCl form how much mass of Na₂SO₄?
<u><em>mass of Na₂SO₄= 26.72 grams</em></u>
Then, 26.72 grams of Na₂SO₄ are formed when 22 grams of NaOH reacts with 21 grams of H₂SO₄. The correct answer is first option.
Learn more about the reaction stoichiometry:
You might be interested in
Answer:
a) volume of ammonium iodide required =349 mL
b) the moles of lead iodide formed = 0.0436 mol
Explanation:
The reaction is:
It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.
Let us calculate the moles of lead nitrate taken in the solution.
Moles=molarityX volume (L)
Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol
the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol
The volume of ammonium iodide required will be:
the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol
Answer:
D) HCl(aq)
Explanation:
A homogeneous mixture can be defined as any liquid, solid or gaseous mixture which has an identical or uniform composition and properties throughout any given sample of the mixture. In Chemistry, all solutions are considered to be a homogeneous mixture.
In this scenario, the chemical formula which represents a homogeneous mixture is aqueous hydrogen chloride, HCl(aq). The aqueous hydrogen chloride is a homogeneous mixture of water and hydrogen chloride. This ultimately implies that, aqueous hydrogen chloride HCl(aq) is a solution of hydrogen chloride in water and it is commonly referred to as Hydrochloric acid.
Given by the chemical equation;