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kolezko [41]
3 years ago
7

What volume of 0.117 M HCl is needed to neutralize 28.67ml of 0.137 m KOH?

Chemistry
1 answer:
shutvik [7]3 years ago
3 0
To determine the volume of the acid needed in the reaction, we first have to know the reaction involved in the process. Since it is a neutralization reaction, then it should produce a salt and water. The reaction would be:

HCl + KOH = KCl + H2O

We use this reaction to relate the substances. We calculate as follows:

0.137 M KOH (.02867 L solution ) = 3.92779x10^-3 mol KOH
3.92779x10^-3 mol KOH ( 1 mol HCl / 1 mol KOH ) = 3.92779x10^-3 mol HCl

Volume of HCl = 3.92779x10^-3 mol HCl / 0.117 M HCl 
                           = 0.03357 L HCl or 33.57 mL HCl needed
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Explanation:

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.950 L of .420 M H2SO4 is mixed with .900 L of .260 M KOH. What concentration of sulfuric acid remains after neutralization?
kipiarov [429]
H₂SO₄:

V=0,95L
Cm=0,420mol/L

n = CmV = 0,42mol/L * 0,95L = 0,399mol

KOH:

V=0,9L
Cm=0,26mol/L

n = CmV = 0,26mol/L * 0,9L = 0,234mol

H₂SO₄            +           2KOH ⇒ K₂SO₄ + 2H₂O
1mol                :           2mol
0,399mol         :           0,234mol
                                    limiting reagent
reamins: 0,399mol - 0,117mol = 0,282mol

n = 0,282mol
V = 0,950L + 0,900L = 1,85L

Cm = n / V = 0,282mol / 1,85L ≈ 0,152M
3 0
3 years ago
Which experiment lead to the discovery that the tiny, dense nucleus contained most of
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Is thiamine mononitrate an ionic bond or covalent bond?
timofeeve [1]
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7 0
3 years ago
Read 2 more answers
0.50 mol A, 0.60 mol B, and 0.90 mol C are reacted according to the following reaction
algol [13]

Reactant C is the limiting reactant in this scenario.

Explanation:

The reactant in the balanced chemical reaction which gives the smaller amount or moles of product is the limiting reagent.

Balanced chemical reaction is:

A + 2B + 3C → 2D + E

number of moles

A = 0.50 mole

B = 0.60 moles

C = 0.90 moles

Taking A as the reactant

1 mole of A reacted to form 2 moles of D

0.50 moles of A will produce \frac{2}{1} = \frac{x}{0.50}

thus 0.50 moles of A will produce 1 mole of D

Taking B as the reactant

2 moles of B reacted to form 2 moles of D

0.60 moles of B reacted to form x moles of D

\frac{2}{2} = \frac{x}{0.6}

x = 2 moles of D is produced.

Taking C as the reactant:

3 moles of C reacted to form 2 moles of D

O.9 moles of C reacted to form x moles of D

\frac{2}{3} = \frac{x}{0.9}

= 0.60 moles of D is formed.

Thus C is the limiting reagent in the given reaction as it produces smallest mass of product.

5 0
3 years ago
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