Chemistry, 05.07.2020 14:01 j015

Cl2 + F2 → ClF3, 5. How many moles of Cl2 are needed to react with 3.44 moles of F2? 6. How many grams of ClF3 form when 0.204 moles of F2 react with excess Cl2? 7. How many grams of ClF3 form from 130.0 grams of Cl2 when F2 is in excess? 8. How many grams of F2 are needed to react with 3.50 grams of Cl2?

Answers: 3

Show answers

Answers

Answer from: mariana2006

5) 1.147 moles Cl2

6) 12.57 grams ClF3

7) 339.10 grams ClF3

8) 5.63 grams F2

Explanation:

Step 1: Data given

Number of moles F2 = 3.44 moles

Molar mass F2 = 38.00 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles F2

For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3

For 3.44 moles F2 we'll need 3.44/3 = 1.147 moles Cl2

Step 1: Data given

Number of moles F2 = 0.204 moles

Molar mass F2 = 38.00 g/mol

Molar mass ClF3 = 92.448 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles ClF3

For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3

For 0.204 moles F2 we'll have 2/3 * 0.204 = 0.136 moles

Step 4: Calculate mass ClF3

Mass ClF3 = Moles ClF3 * molar mass ClF3

Mass ClF3 = 0.136 moles * 92.448 g/mol

Mass ClF3 = 12.57 grams ClF3

Step 1: Data given

Mass of Cl2 = 130.0 grams

Molar mass F2 = 38.00 g/mol

Molar mass ClF3 = 92.448 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles Cl2

Moles Cl2 = mass Cl2 / molar mass Cl2

Moles Cl2 = 130.0 grams / 70.9 g/mol

Moles Cl2 = 1.834 moles

Step 4: Calculate moles

For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3

For 1.834 moles Cl2 e'll have 2*1.834 = 3.668 moles ClF3

Step 5: Calculate mass ClF3

Mass ClF3 = Moles ClF3 * molar mass ClF3

Mass ClF3 = 3.668 moles * 92.448 g/mol

Mass ClF3 = 339.10 grams ClF3

Step 1: Data given

Mass of Cl2 = 3.50 grams

Molar mass F2 = 38.00 g/mol

Molar mass ClF3 = 92.448 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles Cl2

Moles Cl2 = Mass Cl2 / molar mass Cl2

Moles Cl2 = 3.50 grams / 70.9 g/mol

Moles Cl2 = 0.0494 moles

Step 4: Calculate moles F2

For 1 mol Cl2 we need 3 moles F2

For 0.0494 moles we need 3*0.0494 = 0.1482 moles

Step 5: Calculate mass F2

Mass F2 = moles F2 * molar mass F2

Mass F2 = 0.1482 moles * 38.00 g/mol

Mass F2 = 5.63 grams F2

Answer from: Quest

distance, luminosity, brightness, radius, chemical composition, and temperature

Answer from: Quest

there is none because earth is stationary and flat.

Another question on Chemistry

Chemistry, 22.06.2019 01:30

(apex) when a cup of water is dropped, as the cup falls, the water in the cup falls out true or false?

Answers: 1

Answer

Chemistry, 22.06.2019 12:00

Marcel just purchased 1.69 grams of iron fillings in order to make living putty for his 6 year old niece. how many moles of iron are made in his sample?

Answers: 1

Answer

Chemistry, 22.06.2019 17:30

A650 ml sodium bromine solution has a bromide ion concentration of 0.245 m. what is the mass (g) of sodium bromide in solution? a) 103.b)0.00155.c)16400.d) 16.4.e) 0.159

Answers: 2

Answer

Chemistry, 22.06.2019 19:00

Imagine that a new planet is discovered with two moons of equal mass: moon a and moon b. the mass of the new planet is greater than the combined mass of its moons. moon a is farther away from the new planet than moon b. what is the planet's gravitational pull on moon a compared to the planet's gravitational pull on moon b? the planet's gravity repels moon a with a greater force than it repels moon b, which is why moon a is farther away. the gravitational pull on moon b is greater than on moon a because moon b is closer to the new planet than moon a. the gravitational pull on moon b is greater than on moon a because moon b is farther away from the new planet than moon a. the gravitational pull on moon a is the same as the gravitational pull on moon b because distance does not affect the planet's gravity.

Answers: 1

Answer

You know the right answer?

Cl2 + F2 → ClF3, 5. How many moles of Cl2 are needed to react with 3.44 moles of F2? 6. How many gra...

Questions