200ml of 0.05M magnesium chloride is mixed with 75ml of 0.1M silver nitrate solution. find the number of moles and mass in grams of AgCl formed and what is the limiting reagent
Answers
Answer:
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Explanation:
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In the given reaction 0.0075 moles of AgCl are formed having a mass equal to 1.08 grams. AgNO₃ is the limiting reagent for the given reaction.
Given:
Molarity of Magnesium Chloride solution (M₁) = 0.05 M
The volume of Magnesium Chloride solution (V₁) = 200 ml = 0.2 Litre
Molarity of Silver Nitrate solution (M₂) = 0.1 M
The volume of Silver Nitrate solution (V₂) = 75 ml = 0.075 Litre
To Find:
The number of moles and mass of Silver Chloride (AgCl) formed.
The limiting reagent among Magnesium Chloride and Silver Nitrate.
Solution:
→ The molarity of a solution is equal to the number of moles of solute present per liter of the solution.
So by the concept of Moles and Molarity, it can be said that:
Number of Moles = Molarity × Volume
- No. of Moles of MgCl₂ = 0.2 × 0.05 = 0.01 moles
- No. of Moles of AgNO₃ = 0.075 × 0.1 = 0.0075 moles
The reaction between Magnesium Chloride and Silver Nitrate occurs as follows:
MgCl₂ + 2AgNO₃ → Mg(NO₃)₂ + 2AgCl
1 Mole 2 Mole 1 Mole 2 Mole
→ By stoichiometry we know that for 1 mole of MgCl₂, 2 moles of AgNO₃ are required.
→ Therefore for 0.01 moles of MgCl₂, 0.02 moles of AgNO₃ are required.
∵ Only 0.0075 moles of AgNO₃ are present.
∴ AgNO₃ will be the limiting reagent for the above reaction.
→ Since 2 Moles of AgCl are formed from 2 Moles of AgNO₃
∴ 0.0075 Moles of AgCl will be formed from 0.0075 Moles of AgNO₃
MgCl₂ + 2AgNO₃ → Mg(NO₃)₂ + 2AgCl
No. Of Moles: 0.0075 0.0075
→ Molar mass of AgCl = 143.5 gram
∴ 0.0075 moles of AgCl weighs = 143.5 × 0.0075 = 1.08 gram
Hence 0.0075 moles of AgCl are formed having a mass equal to 1.08 grams. AgNO₃ is the limiting reagent for the given reaction.
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