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kicyunya [14]
3 years ago
11

Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. neutraliz

ation reaction: Suppose 0.250 L of 0.450 M H2SO4 is mixed with 0.200 L of 0.240 M KOH . What concentration of sulfuric acid remains after neutralization
Chemistry
1 answer:
madreJ [45]3 years ago
4 0

Answer:

0.197 M

Explanation:

The reaction equation is:

H2SO4(aq) +2KOH(aq) ----> K2SO4(aq) + 2H2O(l)

number of moles of H2SO4 = 0.25 L * 0.45 M = 0.1125 moles

number of moles of KOH = 0.2 L * 0.24 M = 0.048 moles

since H2SO4 is the reactant in excess;

2 moles of KOH reacts with 1 mole of H2SO4

0.048 moles of KOH reacts with 0.048 * 1/2 = 0.024 moles of H2SO4

Amount of excess H2SO4 left unreacted = 0.1125 - 0.024 = 0.0885 moles

Total volume = 0.25 L+ 0.2 L = 0.45 L

concentration of H2SO4 = 0.0885/0.45 = 0.197 M

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If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i
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Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

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If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

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[isobutane] = 2.86 M

Explanation:

Butane  ⇄  Isobutane

At equilibrium

1.0 M               2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M               -

After equilibrium reestablishes:

(1.50-x)M            (2.5+x)

The equilibrium expression will wriiten as:

K_c=\frac{[Isobutane]}{[Butane]}

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