Some additional information is required to answer this question: the volumes of the solutions. So this answer assume 10.0 ml of each solution.
Answer:
- <u>Limiting reagent: crystal violet</u>
- <u>How much excess reagent remains: 0.00099 mol of sodium hydroxide.</u>
Explanation:
This question deals with the first part of a common experiment carried out in a lab, where a solution is preperared by mixing certain volume of crystal violet of known concentration with a volume of sodium hydroxide of known concentration, to obtain information about the kinetics of the reaction.
Both volumes must be known too.
To solve the answer you must know the volumes of each solution.
For explanation purposes, I will assume that the solution is prepared by mixing <u><em>10.0 mL of the 1.0× 10⁻⁵ M crystal viole</em></u>t with <em><u>10.0 mL of the 0.10 M sodium hydroxide.</u></em>
Crystal violet is the organic compound with unit formula formula: C₂₅N₃H₃₀Cl.
<u>1) Chemical equation: </u>
Crystal violet is an ionic compound and the reaction with sodium hydroxied may be represented as folllows:
- C₂₅N₃H₃₀Cl + NaOH → C₂₅N₃H₃₀OH + NaCl
<u>2) Mole ratio:</u>
- 1 mol C₂₅N₃H₃₀Cl : 1 mol NaOH : 1 mol C₂₅N₃H₃₀OH : 1 mol NaCl
So by each mol of crystal violet 1 mol of sodium hydroxide react.
<u>3) Calculate the number of moles of each reactant:</u>
- Molarity: M = n / V (in liters) ⇒ n = M × V
- Crystal violet: n = 10.0 ml × 1 liter / 1,000 ml × 1.00×10 ⁻⁵ M = 1.00×10⁻⁷ mol
- Sodium hydroxide: 10.0 ml × 1 liter / 1,000 ml × 0.1 M = 1.00×10⁻³ mol
<u>3) Limiting reactant:</u>
- Since, as per the theoretical mole ratio, 1.00×10⁻⁵ mol of crystal violet will react with the same number of moles of sodium hydroxide, the former is the limiting reactant.
<u>4) Excess reagent:</u>
The excess reagent is sodium chloride. The amount of excess reagent that remains after the reaction runs to completion is calculated by:
- Remain reagent = Initial amount - amount that reacted
- Since the mole ratio is 1 : 1, 1.00×10⁻⁵ mol of crystal violet react with 1.00×10⁻⁵ mol of sodium hydroxide, and the amount of sodium hydroxide that remains after the reaction runs to completion is:
1.00×10⁻³ mol - 1.00×10⁻⁵ mol = 0.00100 mol - 0.0000100 mol = 0.00099 mol (sodium hydroxide)
