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cluponka [151]
3 years ago
14

How many moles of oxygen are formed when 58.6 g of KNO3 decomposes according to the following reaction? 4 KNO3(s) → 2 K2O(s) + 2

N2(g) + 5 O2(g)
Chemistry
1 answer:
Leya [2.2K]3 years ago
3 0

Answer:

0.725 mol

Explanation:

Moles are calculated as the given mass divided by the molecular mass.

i.e. ,

moles = ( mass / molecular mass )

since,

mass of KNO₃ = 58.6 g  ( given )

Molecular mass of KNO₃ = 101 g / mol

Therefore,

moles of KNO₃ = 58.6 g / 101 g / mol

moles of KNO₃ = 0.58 mol

From the balanced reaction ,

4 KNO₃ (s) ---> 2K₂O (s) + 2N₂ (g) + 5O₂ (g)

By the decomposition of 4 mol of KNO₃ , 5 mol of O₂ are formed ,

hence, unitary method is used as,

1  mol of KNO₃  gives 5 / 4 mol O₂

Therefore,

0.58 mol of KNO₃ , gives , 5 / 4  * 0.58 mol of O₂

Solving,

0.58 mol of KNO₃ , gives , 0.725 mol of O₂

Therefore,

58.6g of KNO₃ gives 0.725 mol of O₂.

You might be interested in
SnO2 + 2 H2 ——> Sn + 2 H2O
SpyIntel [72]

Answer:

0.15g

Explanation:

Given parameters:

Number of molecules of water = 1.2 x 10²¹ molecules

Unknown:

Mass of SnO₂  = ?

Solution:

To solve this problem, we have to work from the known to the unknown specie;

             SnO₂   +    2H₂    →   Sn  +   2H₂O

Ensure that the equation given is balanced;

       

Now,

          the known species is water;

                  6.02 x 10²³ molecules of water  = 1 mole

                   1.2 x 10²¹ molecules of water  = \frac{1.2 x 10^{21} }{6.02 x 10^{23} }    = 0.2 x 10⁻²moles

Number of moles of water  = 0.002moles

           From the balanced chemical equation:

         

             2 mole of water is produced from 1 mole of    SnO₂  

           0.002 moles of water will be produced from \frac{0.002}{2}  = 0.001moles

To find the mass;

           Mass  = number of moles x molar mass

Molar mass of  SnO₂ = 118.7 + 2(16) = 150.7g/mol

        Mass  =  0.001 x 150.7 = 0.15g

3 0
3 years ago
YOU DO:
mihalych1998 [28]

Answer:

67.1%

Explanation:

Based on the chemical equation, if we determine the moles of sodium carbonate, we can find the moles of NaHCO₃ that reacted and its mass, thus:

<em>Moles Na₂CO₃ - 105.99g/mol-:</em>

6.35g * (1mol / 105.99g) = 0.0599 moles of Na₂CO₃ are produced.

As 1 mole of sodium carbonate is produced when 2 moles of NaHCO₃ reacted, moles of NaHCO₃ that reacted are:

0.0599 moles of Na₂CO₃ * (2 moles NaHCO₃ / 1 mole Na₂CO₃) = 0.1198 moles of NaHCO₃

And the mass of NaHCO₃ in the sample (Molar mass: 84g/mol):

0.1198 moles of NaHCO₃ * (84g / mol) = 10.06g of NaHCO₃ were in the original sample.

And percent of NaHCO₃ in the sample is:

10.06g NaHCO₃ / 15g Sample * 100 =

<h3>67.1%</h3>
7 0
3 years ago
Activity 2 Directions: Select among the choices thbest type of material to be used in making the objects at the left and explain
joja [24]

Answer:

A Walk in the City

Make a list of the number of cars, jeeps, tricycle, and even trucks you've seen on your way to school. On a Decision Making Chart, answer the question, "Does the volume of traffic affect the air quality in my local community?" Write your reasons for saying YES on the Reasons for column, and the reasons for saying NO on the Reasons Against column. At the bottom of a chart, make a position by writing your decision on the same question.

Decision Making Chart

[question]

[reason for]

[reason against]

[my decision]

3 0
2 years ago
"acid is responsible for the odor in rancid butter. a solution of 0.25 m butyric acid has a ph of 2.71. what is the ka for"
Salsk061 [2.6K]

Answer:- The Ka for the acid is 1.53*10^-^5 .

Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

C -X +X +X

E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

Ka = [H^+][A^-]\frac{1}{HA}

Where, Ka is the acid ionization constant. Let's plug in the values.

Ka = \frac{X^2}{0.25-X}

Let's calculate the value of X first using the equation:

pH = -log[H^+][/tex]

on taking antilog ob above equation we get:

[H^+]=10^-^p^H

[H^+]=10^-^2^.^7^1

[H^+] = 0.00195

So, X = 0.001195

Let's plug in this value of X in the equation:-

Ka=\frac{(0.00195)^2}{0.25-0.00195}

Ka=1.53*10^-^5

So, the value of Ka for butyric acid is 1.53*10^-^5 .

8 0
3 years ago
Read 2 more answers
Please help me on this!!
Fudgin [204]

Answer:

OK

Explanation:

OK

7 0
3 years ago
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