Respuesta :
Taking into account the definition of molarity and the stoichiometry of the reaction, 0.93336 grams of solid barium sulfate form when 25.0 mL of 0.160 M barium chloride reacts with 68.0 mL of 0.055 M sodium sulfate.
The balanced reaction is:
BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2 NaCl(aq)
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:
- BaCl₂: 1 mole
- Na₂SO₄: 1 mole
- BaSO₄: 1 mole
- NaCl: 2 moles
On the other side, molarity is the number of moles of solute that are dissolved in a given 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, 25.0 mL= 0.025 L (being 1000 mL= 1 L) of 0.160 M 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{number of moles of solute}{0.025 L}[/tex]
Solving:
number of moles of barium chloride= 0.160 M * 0.025 L
number of moles of barium chloride= 0.004 moles
On the other side, 68.0 mL= 0.068 L of 0.055 M sodium sulfate reacts. So, by definition of molarity, the number of moles that participate in the reaction is calculated as:
[tex]0.055 M=\frac{number of moles of solute}{0.068 L}[/tex]
Solving:
number of moles of sodium sulfate= 0.055 M * 0.068 L
number of moles of sodium sulfate= 0.00374 moles
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.
To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction and the amount of moles calculated previously. You can use a simple rule of three as follows: if by stoichiometry 1 mole of barium chloride reacts with 1 mole of sodium sulfate, 0.004 moles of barium chloride reacts how much moles of sodium sulfate?
[tex]amount of moles of sodium sulfate=\frac{0.004 moles of barium chloridex 1mole of sodium sulfate}{1 mole of barium chloride}[/tex]
amount of moles of sodium sulfate= 0.004 moles
But 0.004 moles of sodium sulfate are not available, 0.00374 moles are available. Since you have less moles than you need to react with 0.004 moles of barium chloride, sodium sulfate will be the limiting reagent.
Then, it is possible to determine the amount of moles of barium sulfate produced by another rule of three: if by stoichiometry 1 mole of sodium sulfate produce 1 mole of barium sulfate, 0.0004 moles of sodium sulfate produces how many moles of barium sulfate?
[tex]amount of moles of barium sulfate=\frac{0.0004 moles of sodium sulfatex1 mole of barium sulfate}{1 mole of sodium sulfate}[/tex]
amount of moles of barium sulfate= 0.004 moles
Being the mass molar of barium sulfate 233.34 g/mole, the mass produced of the compound is calculated as:
[tex]0.004 molesx\frac{233.34 grams}{1 mole} =[/tex] 0.93336 grams
Finally, 0.93336 grams of solid barium sulfate form when 25.0 mL of 0.160 M barium chloride reacts with 68.0 mL of 0.055 M sodium sulfate.
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