a. AlCl₃ ⇒ limiting reactant(smaller ratio)
Cu ⇒ excess reactant
b. the mass of leftover reactant : 7.207 g
<h3>Further explanation</h3>
Given
25 g Cu
25 g AlCl3
Required
a. the excess and limiting reactants
b. the mass of leftover reactant
Solution
Reaction
3Cu + 2AlCl₃ ⇒ 3CuCl₂ + 2Al
mol Cu(Ar = 63.5 g/mol) :
mol = mass : Mw
mol = 25 : 63.5
mol = 0.394
mol AlCl3(MW=133,34 g/mol) :
mol = 25 : 133,34 g/mol
mol = 0.187
mol ratio to reaction coefficient Cu : AlCl₃ =
AlCl₃ ⇒ limiting reactant(smaller ratio)
Cu ⇒ excess reactant
b. the mass of leftover reactant :
mol Cu = 3/2 x 0.187 = 0.2805
mol left = 0.394 - 0.2805 = 0.1135
mass = 0.1135 x 63.5 = 7.207 g
Stirring affects how quickly a solute dissolves in a solvent, but has no effect on how much solute will dissolve. The amount of solute that will dissolve is affected by temperature - more will dissolve at higher temperatures. This is called the solubility of the solute
Answer is: 153.52 grams of hypobromous acid <span>must be added.
</span>Chemical dissociation: HBrO ⇄ H⁺ + BrO⁻.
pH = 4.25.
pH = -log[H⁺].
[H⁺] = 10∧(-pH).
[H⁺] = 10∧(-4.25).
[H⁺] = [BrO⁻] = 5.62·10⁻⁵ M.
Ka = [H⁺] · [BrO⁻] / [HBrO].
2.00·10⁻⁹ = (5.62·10⁻⁵ M)² / [HBrO].
[HBrO] = 3.16·10⁻⁹ M² / 2.00·10⁻⁹.
[HBrO] = 1.58 M.
m(HBrO) = n(HBrO) · M(HBrO).
m(HBrO) = 1.58 mol · 96.91 g/mol.
m(HBrO) = 153.52 g.
Answer:
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