Respuesta :
Considering the reaction's stoichiometry and the notion of molarity, 1.167 grams of solid barium sulfate form when 32.0 mL of 0.160 M barium chloride reacts with 70.0 mL of 0.065 M sodium sulfate.
The balanced reaction is:
BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2 NaCl(aq)
What is a Limiting Reagent?
The reactant that is consumed first in a chemical reaction is the limiting reagent because it stops any more reactions from taking place. The limiting reagent controls how much product is produced during the reaction.
Utilizing the reaction stoichiometry and the estimated number of moles, it is possible to identify the limiting reagent.
According to reaction stoichiometry, which describes the proportions of reagents and products in a chemical reaction, the following numbers of moles of each component are involved in the reaction:
- BaCl₂: 1 mole
- Na₂SO₄: 1 mole
- BaSO₄: 1 mole
- NaCl: 2 moles
Molarity, on the other hand, refers to how many moles of solute are dissolved in a certain volume.
Molarity is determined by the expression:
[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]
Molarity is expressed in units[tex]\frac{moles}{L}[/tex].
In this case, 32.0 mL= 0.032L (being 1000 mL= 1 L) of barium chloride reacts. So, by definition of molarity, the number of moles that participate in the reaction is calculated as:
[tex]0.160 M=\frac{No. of moles of solute}{0.032L}[/tex]
No of moles of barium chloride(solute) = 0.160M × 0.032L
No of moles of barium chloride(solute) = 0.00512moles
On the other side,70.0mL= 0.070 L of 0.065 M sodium sulfate reacts. So, by definition of molarity, the number of moles that participate in the reaction is calculated as:
[tex]0.065 M =\frac{No of moles of solute}{0.070L}[/tex]
No of moles of Sodium sulfate= 0.065M × 0.070L
No of moles of Sodium sulfate = 0.0045moles
if by stoichiometry 1 mole of barium chloride reacts with 1 mole of sodium sulfate, 0.00512moles of barium sulfate reacts with how many moles of sodium sulfate?
No. of moles of sodium sulfate=
[tex]\frac{ 0.00512moles of barium sulfate * 1 mole of sodium sulfate}{1 mole of barium chloride}[/tex]
amount of moles of sodium sulfate= 0.00512 moles
But 0.00512moles of sodium sulfate are not available, 0.0045moles moles are available. Since you have fewer moles than you need to react with 0.00512moles of barium chloride, sodium sulfate will be the limiting reagent.
Then, it is possible to determine the number of moles of barium sulfate produced by another rule of three: if by stoichiometry 1 mole of sodium sulfate produces 1 mole of barium sulfate, 0.00512 moles of sodium sulfate produces how many moles of barium sulfate?
[tex]Amt of mole of barium sulfate=\frac{0.00512 moles of sodium sulfate* 1mole of barium sulfate}{1 mole of sodium sulfate}[/tex]
amount of moles of barium sulfate= 0
Being the mass molar of barium sulfate is 233.34 g/mole, the mass-produced of the compound is calculated as:
= [tex]\frac{0.00512 moles * 233.34 gm}{1 mole}[/tex]
=1.167 gm
Hence, 1.167 gm of solid barium sulfate is formed.
To know more about stoichiometry refer to:-https://brainly.com/question/14935523
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