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Lunna [17]
4 years ago
9

A 7.11g sample of potassium chlorate was decomposed according to the following equation: 2KCLO3 -->2KCL + 3O2 How many moles

of oxygen are formed?
Chemistry
1 answer:
Ivan4 years ago
3 0

Answer:

The answer to your question is 0.087 moles of Oxygen

Explanation:

Data

mass of KClO₃ = 7.11 g

moles of O₂ = ?

Balanced chemical reaction

               2KClO₃  ⇒   2KCl  +  3O₂

Process

1.- Calculate the molar mass of KClO₃

KClO₃ = (1 x 39.1) + (1 x 35.45) + (3 x 16)

           = 39.1 + 35.45 + 48

           = 122.55 g

2.- Convert the mass to KClO₃ to moles

                  122.55 g ------------------- 1 mol

                      7.11 g   -------------------  x

                       x = (7.11 x 1)/122.55

                       x = 0.058 moles

3.- Calculate the moles of Oxygen

                  2 moles of KClO₃ ---------------- 3 moles of Oxygen

                  0.058 moles         ----------------  x

                  x = (0.058 x 3) / 2

                  x = 0.087 moles of Oxygen

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Choose the covalent compounds from the following choices Br2 MgS SO2 KF
Novosadov [1.4K]

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

 

                Less than 0.4 then it is Non Polar Covalent Bond

                

                Between 0.4 and 1.7 then it is Polar Covalent Bond

            

                Greater than 1.7 then it is Ionic

 

For Br₂;

                    E.N of Bromine      =   2.96

                    E.N of Bromine      =   2.96

                                                   ________

                    E.N Difference             0.00          (Non Polar Covalent Bond)


For MgS;

                    E.N of Sulfur               =   2.58

                    E.N of Magnesium      =   1.31

                                                   ________

                    E.N Difference                  1.27          (Ionic Bond)


For SO₂;

                    E.N of Oxygen      =   3.44

                    E.N of Sulfur          =   2.58

                                                   ________

                    E.N Difference             0.86          (Polar Covalent Bond)


For KF;

                    E.N of Fluorine          =   3.98

                    E.N of Potassium      =   0.82

                                                   ________

                    E.N Difference                3.16          (Ionic Bond)

Result: The Bonds in Br₂ and SO₂ are Covalent in Nature.

3 0
3 years ago
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Answer:

carbon, nitrogen,oxygen,flourine

3 0
3 years ago
water (H2O) and carbon dioxide (CO2) are similar because they both contain at least one oxygen A.) atom. B.) mixture C.) molecul
vova2212 [387]

Answer:

a

Explanation:

it is A because h20 and c02 are molecules while what make them up are atoms

5 0
3 years ago
Read 2 more answers
A gas has a density of 1.57 g/L at 40.0 °C and 2.00 atm of pressure. What is the identity of the gas?
Naddika [18.5K]

Answer:

Neon

Explanation:

Step 1: Given and required data

  • Density of the gas (ρ): 1.57 g/L
  • Temperature (T): 40.0°C
  • Pressure (P): 2.00 atm
  • Ideal gas constant (R): 0.08206 atm.L/mol.K

Step 2: Convert T to Kelvin

We will use the following expression.

K = °C + 273.15 = 40.0 + 273.15 = 313.2 K

Step 3: Calculate the molar mass of the gas (M)

For an ideal gas, we will use the following expression.

ρ = P × M/R × T

M = ρ × R × T/P

M = 1.57 g/L × 0.08206 atm.L/mol.K × 313.2 K/2.00 atm

M = 20.17 g/mol

The gas with a molar mass of 20.17 g/mol is Neon.

6 0
3 years ago
Problem Page Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.2 g o
stich3 [128]

Answer: 0.0 grams

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of butane

\text{Number of moles}=\frac{5.2g}{58.12g/mol}=0.09moles

b) moles of oxygen

\text{Number of moles}=\frac{32.6g}{32g/mol}=1.02moles

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According to stoichiometry :

2 moles of butane require 13 moles of O_2

Thus 0.09 moles of butane will require =\frac{13}{2}\times 0.09=0.585moles  of O_2

Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.

Thus all the butane will be consumed and 0.0 grams of butane will be left.

7 0
3 years ago
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