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MaRussiya [10]
3 years ago
14

A solution is made by dissolving

Chemistry
1 answer:
barxatty [35]3 years ago
8 0

Answer:

4.52 mol/kg

Explanation:

Given data:

Mass of lithium fluoride = 22.1 g

Mass of water = 188 g

Molality = ?

Solution:

Molality:

It is the number of moles of solute into kilogram of solvent.

Formula:

Molality = number of moles of solute / kilogram solvent

Mathematical expression:

m = n/kg

Now we will convert the grams of LiF into moles.

Number of moles = mass/ molar mass

Number of moles = 22.1 g/ 26 g/mol

Number of moles = 0.85 mol

Now we will convert the g of water into kg.

Mass of water = 188 g× 1kg/1000 g = 0.188 kg

Now we will put  the values in formula.

m = 0.85 mol / 0.188 kg

m = 4.52 mol/kg

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The balanced chemical reaction:

K2SO4 + O2 = 2KO2 + SO2

Assuming that the reaction is complete, all of the potassium sulfate is consumed. We relate the substances using the chemical reaction. We calculate as follows:

7.20 g K2SO4 ( 1 mol / 174.26 g) ( 1 mol O2 / 1 mol K2SO4 ) ( 32 g / 1 mol ) = 1.32 g O2 consumed in the reaction.
5 0
3 years ago
Consider the following chemical reaction: 2KCl + 3O2 --> 2KClO3. If you are given 100.0 moles of KCl and 100.0 moles of O2...
g100num [7]

Answer:

O₂; KCl; 33.3  

Explanation:

We are given the moles of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble all the data in one place.

            2KCl  +  3O₂ ⟶ 2KClO₃

n/mol:  100.0   100.0

1. Identify the limiting reactant

(a) Calculate the moles of KClO₃ that can be formed from each reactant

(i)From KCl

\text{Moles of KClO}_{3} = \text{100.0 mol KCl} \times \dfrac{\text{2 mol KClO}_{3}}{\text{2 mol KCl}} = \text{100.0 mol KClO}_{3}

(ii) From O₂

\text{Moles of KClO}_{3} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KClO}_{3}}{\text{3 mol O}_{2}} = \text{66.67 mol KClO}_{3}

O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.

KClO₃ is the excess reactant.

2. Moles of KCl left over

(a) Moles of KCl used

\text{Moles used} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KCl}}{\text{3 mol O}_{2}} = \text{66.67 mol KCl}

(b) Moles of KCl left over

n = 100.0 mol - 66.67 mol = 33.3 mol

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3 years ago
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Calculating speed or atoms and molecules or planets!




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Which formula represents an ionic bond 
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Explanation:

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