Respuesta :
Answer:
a) HClO4 Volume= 38.04mL
b) HCl Volume= 0.19 litters = 190mL
c) AgNO3 Molarity= 0.41 mol/l = 0.41 M
d) KOHmass = 0.275g
Explanation:
In both, the neutralization reactions (a, b and d) and in the single exchange reactions (d) the relationship between the moles of the reagents indicates how much is needed to complete the reaction.
The number of moles of a substance can be calculated as:
moles= Molarity* volume (liters) >For solutions
or
moles= mass(g)/MM (g/mol) > For pure substance
MM= Molecular Mass
a) moles of NaOH = moles of HClO4
NaOH Molarity*NaOH Volume= HClO4 Molarity* HClO4 Volume
HClO4 Volume= (NaOH Molarity* NaOH Volume)/HClO4 Molarity
HClO4 Volume= (0.0875M*50.00mL)/0.115M
HClO4 Volume= 38.04mL
b) moles of Mg(OH)2 =2 moles of HCl
Mg(OH)2 mass/ Mg(OH)2 MM = 2*HCl Molarity* HCl Volume
HCl Volume= (Mg(OH)2 mass/ Mg(OH)2 MM )/ 2*HCl Molarity
HCl Volume= (2.87g/58.32g/mol)/ 2*0.128mol/l
HCl Volume= 0.19 l =190ml
c) moles of KCl = moles of AgNO3
KCl mass/ KCl MM = AgNO3 Molarity* AgNO3 Volume
AgNO3 Molarity= (KCl mass/ KCl MM )/AgNO3 Volume
AgNO3 Molarity= (0.785g/74.55 g/mol))/0.0258 l
AgNO3 Molarity= 0.41 mol/l = 0.41 M
d) moles of KOH= moles of HCl
KOHmass/ KOH MM = HCl Molarity* HCl Volume
KOHmass = HCl Molarity* HCl Volume*KOH MM
KOHmass = 0.108 (mol/l) *0.0453 l*56.11( g/mol)
KOHmass = 0.275g
(a) The volume of HClO₄ needed to neutralize the NaOH is 38.04 mL
(b) The volume of HCl needed to neutralize the Mg(OH)₂ is 768.75 mL
(c) The molarity of the AgNO₃ solution is 0.4549 M
(d) The mass of KOH that must be present in the solution is 0.274 g
(a)
First, we will write a balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
HClO₄ + NaOH → NaClO₄ + H₂O
This means
1 mole of HClO₄ is needed to neutralize 1 mole of NaOH
Now, we will determine number of moles of NaOH present
From the question.
Volume of NaOH = 50.0 mL = 0.05 L
Concentration of NaOH = 0.0875 M
Using the formula
Number of moles = Concentration × Volume
∴ Number of moles of NaOH present = 0.0875 × 0.05
Number of moles of NaOH present = 0.004375 moles
The number of moles of NaOH present is 0.004375 moles
Since,
1 mole of HClO₄ is needed to neutralize 1 mole of NaOH
Then,
0.004375 moles of HClO₄ will be needed to neutralize 0.004375 moles mole of NaOH
∴ The number of moles of HClO₄ needed to neutralize the NaOH is 0.004375 moles
Now, for the volume of HClO₄ needed,
Concentration of HClO₄ = 0.115 M
From the formula
[tex]Volume =\frac{Number\ of\ moles }{Concentration}[/tex]
∴ Volume of HClO₄ needed = [tex]\frac{0.004375}{0.115}[/tex]
Volume of HClO₄ needed = 0.03804 L
Volume of HClO₄ needed = 38.04 mL
Hence, the volume of HClO₄ needed to neutralize the NaOH is 38.04 mL
(b)
To determine the volume of HCl needed to neutralize the Mg(OH)₂
First, we will write a balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
2HCl + Mg(OH)₂ → MgCl₂ + 2H₂O
This means 2 moles of HCl is needed to neutralize 1 mole of Mg(OH)₂
Now, we will determine the number of moles of Mg(OH)₂ present
From the question
Mass of Mg(OH)₂ = 2.87 g
Using the formula
[tex]Number \ of \ moles = \frac{Mass}{Molar\ mass}[/tex]
Molar mass of Mg(OH)₂ = 58.32 g/mol
∴ Number of moles of Mg(OH)₂ present = [tex]\frac{2.87}{58.32}[/tex]
Number of moles of Mg(OH)₂ present = 0.0492 moles
From the balanced chemical equation,
2 moles of HCl is needed to neutralize 1 mole of Mg(OH)₂
∴ (2 × 0.0492) moles of HCl will be needed to neutralize 0.0492 moles of Mg(OH)₂
2 × 0.0492 = 0.0984 moles
∴ 0.0984 moles of HCl is needed
Now, for the volume of HCl needed
Concentration of HCl = 0.128 M
From the formula
[tex]Volume =\frac{Number\ of\ moles }{Concentration}[/tex]
∴ Volume of HCl needed = [tex]\frac{0.0984}{0.128}[/tex]
Volume of HCl needed = 0.76875 L
Volume of HCl needed = 768.75 mL
Hence, the volume of HCl needed to neutralize the Mg(OH)₂ is 768.75 mL
(c)
To determine the molarity of the AgNO₃ solution
First, we will write a balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
AgNO₃ + KCl → AgCl + KNO₃
This means, 1 mole of AgNO₃ is required to precipitate all the Cl⁻ ions in 1 mole of KCl
Now, we will determine the number of moles of KCl present
Mass of KCl = 785 mg = 0.785 g
From the formula
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]
Molar mass of KCl = 74.55 g/mol
∴ Number of moles of KCl = [tex]\frac{0.875}{74.55}[/tex]
Number of moles of KCl = 0.011737 moles
Since 1 mole of AgNO₃ is required to precipitate all the Cl⁻ ions in 1 mole of KCl
Then,
0.011737 moles of AgNO₃ will be required to precipitate all the Cl⁻ ions in 0.011737 mole KCl
∴ The number of moles of the AgNO₃ is 0.011737
Now, for the molarity (that is, concentration) of the AgNO₃ solution
From the question
Volume of AgNO₃ = 25.8 mL = 0.0258 L
Using the formula
[tex]Concentration = \frac{Number\ of\ moles}{Volume}[/tex]
∴ Concentration of the AgNO₃ = [tex]\frac{0.011737}{0.0258}[/tex]
Concentration of the AgNO₃ = 0.4549 M
Hence, the molarity of the AgNO₃ solution is 0.4549 M
(d)
To determine the mass of KOH that must be present,
First, we will write a balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
HCl + KOH → KCl + H₂O
This means 1 mole of HCl is needed to neutralize 1 mole of KOH
Now, we will determine the number of moles of HCl present
From the question
Concentration of HCl = 0.108 M
Volume of HCl = 45.3 mL = 0.0453 L
Using the formula
Number of moles = Concentration × Volume
∴ Number of moles of HCl present = 0.108 × 0.0453
Number of moles of HCl present = 0.0048924 moles
Since 1 mole of HCl is needed to neutralize 1 mole of KOH
Then,
0.0048924 moles of HCl will neutralize 0.0048924 moles of KOH
∴ The number of moles of KOH needed is 0.0048924 moles
Now, for the mass of KOH
From the formula
Mass = Number of moles × Molar mass
Molar mass of KOH = 56.1 g/mol
∴ Mass of KOH = 0.0048924 × 56.1
Mass of KOH = 0.27446 g
Mass of KOH ≅ 0.274 g
Hence, the mass of KOH that must be present in the solution is 0.274 g
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