Answer:
9.9 ml of 0.200M NH₄OH(aq)
Explanation:
3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)
?ml of 0.200M NH₄OH(aq) reacts completely with 12ml of 0.550M FeCl₃(aq)
1 x Molarity NH₄OH x Volume Am-OH Solution(L) = 2 x Molarity FeCl₃ x Volume FeCl₃ Solution
1(0.200M)(Vol Am-OH Soln) = 3(0.550M)(0.012L)
=> Vol Am-OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liter = 9.9 milliliters
The amount of [tex]0.200 M NH_{4} OH[/tex] required to react with [tex]12.0 ml[/tex] of [tex]0.550 M FeCl_{3}[/tex]would be [tex]9.9 ml.[/tex]
Given that,
Reaction:
[tex]3NH_{4} OH(Iaq) + FeCl_{3} (aq) => NH_{4} Cl(aq) + Fe(OH)_{3} (s)[/tex]
To find,
The amount of [tex]0.200 M NH_{4} OH[/tex]required to react with [tex]12.0 ml[/tex] of [tex]0.550 M FeCl_{3}[/tex] = ?
Procedure:
As we know,
[tex]1[/tex] × molarity of [tex]NH_{4}OH[/tex]× [tex]Volume (Am-OH) Solution(l)[/tex] [tex]= 2[/tex] × molarity of [tex]FeCl_{3}[/tex]× volume of solution [tex]FeCl_{3}[/tex]
By substituting the given values in this formula, we get
⇒ [tex]1(0.200M)[/tex][tex](Vol Am-OH Sol)= 3(0.550M)(0.012L)[/tex]
⇒ [tex]Vol Am-OH Soln = 3(0.550M)(0.012L)/1(0.200M)[/tex]
⇒ [tex]Vol Am-OH Solution = 0.099 liters[/tex]
[tex]= 0.099[/tex] × [tex]1000[/tex]
[tex]=[/tex] [tex]9.9 milliliters[/tex]
Thus, [tex]9.9 ml[/tex] is the correct answer.
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