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Y_Kistochka [10]
2 years ago
7

When heated, KClO3 decomposes into KCl and O2. 2KClO3⟶2KCl+3O2 If this reaction produced 13.2 g KCl, how many grams of O2 were p

roduced?
Chemistry
1 answer:
MissTica2 years ago
7 0

Answer:

8.5gm O2 produced

Explanation:

When heated, KClO3 decomposes into KCl and O2. 2KClO3⟶2KCl+3O2 If this reaction produced 13.2 g KCl, how many grams of O2 were produced?

for every 2 moles of KCl produced, 3 moles of O2 are produced

Mol weight of KCl =39+35.5=74.5gm

13.2 gm KCl = 13.2/74.5 = 0.177 moles

this will make (3/2) X 0.177 = 0.2655 moles of O2

O2 mol wt is 32  0.2655 X32 = 8.5gm O2 produced

You might be interested in
A mysterious white powder could be powdered sugar (C12H22O11), cocaine (C17H21NO4), codeine (C18H21NO3), norfenefrine (C8H11NO2)
rodikova [14]

Norfenefrine (C₈H₁₁NO₂).

<h3>Further explanation</h3>

We will solve a case related to one of the colligative properties, namely freezing point depression.

The freezing point of the solution is the temperature at which the solution begins to freeze. The difference between the freezing point of the solvent and the freezing point of the solution is called freezing point depression.

\boxed{ \ \Delta T_f = T_f(solvent) - T_f(solution) \ } \rightarrow \boxed{ \ \Delta T_f = K_f \times molality \ }

<u>Given:</u>

A mysterious white powder could be,

  • powdered sugar (C₁₂H₂₂O₁₁) with a molar mass of 342.30 g/moles,
  • cocaine (C₁₇H₂₁NO₄) with a molar mass of 303.35 g/moles,
  • codeine (C₁₈H₂₁NO₃) with a molar mass of 299.36 g/moles,
  • norfenefrine (C₈H₁₁NO₂) with a molar mass of 153.18 g/moles, or
  • fructose (C₆H₁₂O₆) with a molar mass of 180.16 g/moles.

When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.

<u>Question: </u>What is the identity of the white powder?

<u>The Process:</u>

Let us identify the solute, the solvent, initial, and final temperatures.

  • The solute = the powder
  • The solvent = ethanol
  • The freezing point of the solvent = −114.6°C
  • The freezing point of the solution = −115.5°C

Prepare masses of solutes and solvents.

  • Mass of solute = 82 mg = 0.082 g
  • Mass of solvent = density x volume, i.e., \boxed{ \ 0.789 \ \frac{g}{cm^3} \times 1.50 \ cm^3 = 1.1835 \ g = 0.00118 \ kg  \ }

We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.

The molality formula is as follows:

\boxed{ \ m = \frac{moles \ of \ solute}{kg \ of \ solvent} \ } \rightarrow \boxed{ \ m = \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }

Now we combine it with the formula of freezing point depression.

\boxed{ \ \Delta T_f =  K_f \times \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }

It is clear that we will determine the molar mass of the solute (denoted by Mr).

We enter all data into the formula.

\boxed{ \ -114.6^0C - (-115.5^0C) = 1.99 \frac{^0C}{m} \times \frac{0.082 \ g}{Mr \times 0.00118 \ kg} \ }

\boxed{ \ 0.9 = \frac{1.99 \times 0.082}{Mr \times 0.00118} \ }

\boxed{ \ Mr = \frac{0.16318}{0.9 \times 0.00118} \ }

We get \boxed{ \ Mr = 153.65 \ }

These results are very close to the molar mass of norfenefrine which is 153.18 g/mol. Thus the white powder is norfenefrine.

<h3>Learn more</h3>
  1. The molality and mole fraction of water brainly.com/question/10861444
  2. About the mass and density of ethylene glycol as an  antifreeze brainly.com/question/4053884
  3. About the solution as a homogeneous mixture  brainly.com/question/637791

Keywords: a mysterious white powder, sugar, cocaine, codeine, norfenefrine, fructose, the solute, the solvent, dissolved, ethanol, normal freezing point, the freezing point depression, the identity

7 0
3 years ago
Read 2 more answers
How many moles are in 39.5 grams of Lithium?
Blizzard [7]

Answer:

185.05 g.

Explanation

Firstly, It is considered as a stichiometry problem.

From the balanced equation: 2LiCl → 2Li + Cl₂

It is clear that the stichiometry shows that 2.0 moles of LiCl is decomposed to give 2.0 moles of Li metal and 1.0 moles of Cl₂, which means that the molar ratio of LiCl : Li is (1.0 : 1.0) ratio.

We must convert the grams of Li metal (30.3 g) to moles (n = mass/atomic mass), atomic mass of Li = 6.941 g/mole.

n = (30.3 g) / (6.941 g/mole) = 4.365 moles.

Now, we can get the number of moles of LiCl that is needed to produce 4.365 moles of Li metal.

Using cross multiplication:

2.0 moles of LiCl → 2.0 moles of Li, from the stichiometry of the balanced equation.

??? moles of LiCl → 4.365  moles of Li.

The number of moles of LiCl that will produce 4.365 moles of Li (30.3 g) is (2.0 x 4.365 / 2.0) = 4.365 moles.

Finally, we should convert the number of moles of LiCl into grams (n = mass/molar mass).

Molar mass of LiCl = 42.394 g/mole.

mass = n x molar mass = (4.365 x 42.394) = 185.05 g.

7 0
3 years ago
Read 2 more answers
Atoms of metallic elements can form ionic bonds, but they aren't very good at forming covalent bonds. why?
Solnce55 [7]
<span>They have a great tendency to lose electrons</span>
7 0
3 years ago
When 25.0 g of ch4 reacts completely with excess chlorine yielding 45.0 g of ch3cl, what is the percentage yield, according to c
Ber [7]

Taking into account definition of percent yield, the percent yield for the reaction is 57.08%.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CH₄ + Cl₂ → CH₃Cl + HCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • CH₄: 1 mole
  • Cl₂: 1 mole
  • CH₃Cl: 1  mole
  • HCl:  1 mole

The molar mass of the compounds is:

  • CH₄: 16 g/mole
  • Cl₂: 70.9 g/mole
  • CH₃Cl: 50.45 g/mole
  • HCl:  36.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • CH₄: 1 mole ×16 g/mole= 16 grams
  • Cl₂: 1 mole ×70.9 g/mole= 70.9 grams
  • CH₃Cl: 1 mole ×50.45 g/mole= 50.45 grams
  • HCl: 1 mole ×36.45 g/mole= 36.45 grams

Mass of CH₃Cl formed

The following rule of three can be applied: if by reaction stoichiometry 16 grams of CH₄ form 50.45 grams of CH₃Cl, 25 grams of CH₄ form how much mass of CH₃Cl?

mass of CH_{3} Cl=\frac{25 grams of CH_{4}x 50.45grams of CH₃Cl }{16 grams of CH_{4}}

<u><em>mass of CH₃Cl= 78.83 grams</em></u>

Then, 78.83 grams of CH₃Cl can be produced from 25 grams of CH₄.

<h3>Percent yield</h3>

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

percent yield=\frac{actual yield}{theorical yield}x100

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

Percent yield for the reaction in this case

In this case, you know:

  • actual yield= 45 grams
  • theorical yield= 78.83 grams

Replacing in the definition of percent yields:

percent yield=\frac{45 grams}{78.83 grams}x100

Solving:

<u><em>percent yield= 57.08%</em></u>

Finally, the percent yield for the reaction is 57.08%.

Learn more about

the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

percent yield:

brainly.com/question/14408642

#SPJ1

5 0
2 years ago
The concentration of a solution can be changed by _______ solute.
Snezhnost [94]
Add more solute..............
7 0
3 years ago
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