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Katarina [22]
3 years ago
15

Write the balanced neutralization reaction that occurs between H 2 SO 4 and KOH in aqueous solution. Phases are optional. neutra

lization reaction: Suppose 0.150 L of 0.420 M H 2 SO 4 is mixed with 0.100 L of 0.210 M KOH . What concentration of sulfuric acid remains after neutralization
Chemistry
1 answer:
Mariulka [41]3 years ago
3 0

Answer:

0.168 M

Explanation:

First, this is a reaction between the a strong base and a strong acid, therefore, we do not have to count with the acid constant of equilibrium. This reaction is taking place completely and occurs a neutralization, which is the following reaction:

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

Now that we have the reaction, we can go to the second part of the question.

To calculate the remaining concentration after neutralization, we need to calculate the moles of the reactants and determine which is the limiting reactant.

The moles of the reactants:

moles A = 0.42 * 0.15 = 0.063 moles

moles B = 0.210 * 0.1 = 0.021 moles

Now that we have the moles, let's calculate the limiting reactant:

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

If:

1 moles A ---------> 2 moles B

0.063 A -----------> X

X = 0.063 * 2 = 0.126 moles of B

However, we only have 0.021 moles of base, so, this is the limiting reactant.

Now that we know this, let's see the remaining moles of the acid, after the base reacts completely:

moles of A remaining = 0.063 - 0.021 = 0.042 moles

Finally to get the concentration, we have the volume of acid and the base together, so, the final volume is 0.25 L:

C = 0.042 / 0.25 = 0.168 M

This is the final concentration of the acid

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What is the mass of the solid NH4Cl formed when 75.5 g of NH3 is mixed with an equal mass of HCl? What is the volume of the gas
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Answer : The volume of the gas remaining is 56.5 liters.

The gas is hydrochloric acid and the formula of the gas is HCl.

The mass of NH_4Cl produced is, 110.7 grams.

Explanation :

The balanced chemical reaction will be:

NH_3+HCl\rightarrow NH_4Cl

First we have to calculate the moles of NH_3 and HCl

\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}

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\text{Moles of }NH_3=\frac{75.5g}{17g/mole}=4.44mole

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\text{Moles of }HCl=\frac{\text{Mass of }HCl}{\text{Molar mass of }HCl}

Molar mass of HCl = 36.5 g/mole

\text{Moles of }HCl=\frac{75.5g}{36.5g/mole}=2.07mole

Now we have to calculate the limiting and excess reagent.

From the balanced reaction we conclude that

As, 1 mole of HCl react with 1 mole of NH_3

So, 2.07 mole of HCl react with 2.07 mole of NH_3

From this we conclude that, NH_3 is an excess reagent because the given moles are greater than the required moles and HCl is a limiting reagent and it limits the formation of product.

The remaining moles of HCl gas = 4.44 - 2.07 = 2.37 moles

Now we have to calculate the volume of the gas remaining.

Using ideal gas equation :

PV = nRT

where,

P = Pressure of gas = 752 mmHg = 0.989 atm     (1 atm = 760 mmHg)

V = Volume of gas = ?

n = number of moles of gas = 2.37 moles

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 14.0^oC=273+14.0=287K

Putting values in above equation, we get:

0.989atm\times V=2.37mole\times (0.0821L.atm/mol.K)\times 287K

V = 56.5 L

Now we have to calculate the moles of NH_4Cl

As, 1 mole of HCl react with 1 mole of NH_4Cl

So, 2.07 mole of HCl react with 2.07 mole of NH_4Cl

Now we have to calculate the mass of NH_4Cl

\text{ Mass of }NH_4Cl=\text{ Moles of }NH_4Cl\times \text{ Molar mass of }NH_4Cl

Molar mass of NH_4Cl = 53.5 g/mole

\text{ Mass of }NH_4Cl=(2.07moles)\times (53.5g/mole)=110.7g

Thus, the volume of the gas remaining is 56.5 liters.

The gas is hydrochloric acid and the formula of the gas is HCl.

The mass of NH_4Cl produced is, 110.7 grams.

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