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2 years ago

How many moles of water are produced when 3.45 moles of KMnO4 react? Give your answer to the nearest 0.1 moles. moles H2O

Chemistry

1 answer:

andrew11 [14]2 years ago

7 0

Answer:

13.8 moles of water produced.

Explanation:

Given data:

Moles of KMnO₄ = 3.45 mol

Moles of water = ?

Solution:

Chemical equation:

16HCl + 2KMnO₄ → 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O

Moler ratio of water and KMnO₄:

KMnO₄ : H₂O

2 : 8

3.45 : 8/2×3.45 = 13.8 mol

Hence, 3.45 moles of KMnO₄ will produced 13.8 mol of water.

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2 years ago

In the reaction BaCO3 + 2HNO3 + Ba(NO3)2 + CO2 + H2O, what mass of Ba(NO3)2 can be formed by combining 55 g BaCO3 and 26 g HNO3

Nataliya [291]

From the stoichiometry of the reaction, the mass of barium nitrate produced is 54.9 g.

<h3>Stoichiometry</h3>

The term stoichiometry refers to mass - volume relationships. Stoichiometry can be used to calculate the amount, mass or volume of reactants and products from the balanced reaction equation.

The equation of the reaction is written as follows;

BaCO3 + 2HNO3 ------> Ba(NO3)2 + CO2 + H2O

Number of moles of BaCO3 = 55 g/197.34 g/mol = 0.28 moles

Number of moles of HNO3 = 26 g/63.01 g/mol = 0.41 moles

From the reaction equation;

1 mole of BaCO3 reacts with 2 moles of HNO3

0.28 moles of BaCO3 reacts with 0.28 moles × 2 moles/1 mole = 0.56 moles

There is not enough HNO3 hence it is the limiting reactant.

Number of moles of Ba(NO3)2 produced is obtained from;

2 moles of HNO3 yields 1 mole of Ba(NO3)2

0.41 moles of HNO3 yields 0.41 moles × 1 mole/2 moles

= 0.21 moles of Ba(NO3)2

Mass of Ba(NO3)2 = 0.21 moles × 261.337 g/mol = 54.9 g

Learn more about stoichiometry: brainly.com/question/9743981

8 0

2 years ago

Calculate the pH of the following simple solutions:

IceJOKER [234]

Answer :

(1) pH = 1.27

(2) pH = 13.35

(3) The given solution is not a buffer.

Explanation :

Concentration of HCl = $53.1mM=53.1\times 10^{-3}M$

As HCl is a strong acid. So, it dissociates completely to give hydrogen ion and chloride ion.

So, Concentration of hydrogen ion= $53.1\times 10^{-3}M$

pH : It is defined as the negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$pH=-\log (53.1\times 10^{-3})$

$pH=1.27$

Concentration of KOH = 0.223 M

As KOH is a strong base. So, it dissociates completely to give hydroxide ion and potassium ion.

So, Concentration of hydroxide ion= 0.223 M

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (0.223)$

$pOH=0.65$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-0.65=13.35$

Buffer : It is defined as a solution that maintain the pH of the solution by adding the small amount of acid or a base.

It is not a buffer because HCl is a strong acid and KOH is a strong base. Both dissociates completely.

As we know that the pH of strong acid and strong base solution is always 7.

So, the given solution is not a buffer.

5 0

2 years ago