Answer:
The chemical formula of the anion formed =
Explanation:
Acids are the species which furnish hydrogen ions in the solution or is capable of forming bonds with electron pair species as they are electron deficient species.
When an acid donates a proton, it changes into a base which is known as its conjugate base.
The equation for the release of the one hydrogen ion from is shown below as;-
The chemical formula of the anion formed =
<h3>
Answer:</h3>
Limiting reagent: Potassium iodide
Mass of the precipitate (PbI₂) is 4.453 g
<h3>
Explanation:</h3>
We are given;
- 60.0 mL of 0.322 M potassium iodide
- 20.0 mL of 0.530 M lead () nitrate
We are required to identify the limiting reactant and determine the mass of the precipitate formed.
<h3>Step 1: Write the balanced equation for the reaction</h3>
- The balanced equation for the reaction between potassium iodide and lead (II) nitrate is given by;
2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂(s)
<h3>Step 2: Determine the number of moles of the reagents</h3>
Moles of KI
Moles = Molarity × volume
Moles of KI = 0.322 M × 0.060 L
= 0.01932 moles
Moles of KNO₃
Moles = 0.530 M × 0.020 L
= 0.0106 M
From the equation;
- 2 moles of KI reacts with 1 mole of Pb(NO)₂
- Therefore; 0.01932 moles of KI will require 0.00966 moles of Pb(NO₃)₂
- This means, KI is the limiting reagent while Pb(NO₃)₂ is the excess reagent.
<h3>Step 3: Determine the mass of the precipitate PbI₂</h3>
2 moles of KI reacts to produce 1 mole of PbI₂
Therefore;
Moles of PbI₂ = Moles of KI ÷ 2
= 0.01932 moles ÷ 2
= 0.00966 moles
But molar mass of PbI² is 461.01 g/mol
Therefore;
Mass of PbI₂ = 0.00966 moles × 461.01 g/mol
= 4.453 g
Therefore, the mass of the precipitate formed (Pbi₂)is 4.453 g
It's known as chromium (III) sulphate which is green in colour.
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