Equation solving for chemistry

Question

How many grams of KClO3 are needed to form 26.4 g of O2?

2KClO3(s)—>2KCl(s)+3O2(g)

step one show strategy for solving this problem.

1: grams O2–> _______—>_______—>grans KClO

answer bank

moles KClO3

Moles KCl

Grams KCl

Grams KClO3

Moles O2

Grams O2

step 2: show the conversions required to solve this problem and calculate the grams of KClO3.

26.4gO2 X ______/______ X _______/_______ X ______/______ =gKClO3

Answer Bank

122.65 g KClO3

32.00 gO2

74.55 g KCl

1 mole KCl

1 mole O2

1 mole KClO3

2 moles KCl

3 moles O2

2 moles KclO3

grams of KClO3?

Answer 1

step one show strategy for solving this problem.

Answer:

grams O₂ x ( Moles O₂ / Grams O₂ ) x ( moles KClO₃ / Moles O₂ ) x ( Grams KClO₃ / moles KClO₃ ) -> Grams KClO₃

(You don't need KCl to solve)

step 2: show the conversions required to solve this problem and calculate the grams of KClO3.

26.4gO2 X ______/______ X _______/_______ X ______/______ =gKClO3

Answer:

26.4gO₂ X 1 mole O₂/32.00 gO₂ X 2 moles KclO₃/3 moles O₂ X 122.65 g KClO₃/1 mole KClO₃ =gKClO₃

grams of KClO₃? 67.46 g KClO₃

Answer 2

Hi Breanna! For both subparts of this question, we need to use stoichiometry to get from grams O₂ to grams KClO_3. We are provided with the balanced equation for the decomposition reaction: 2KClO₃ (s) --> 2KCl (s) + 3O₂ (g) which tells us how the reactants and products relate to each other in terms of moles: for every 2 moles of KClO₃ that decompose, we get 2 moles of KCl and 3 moles of O₂.

1) grams O₂ --> moles O₂ --> moles KClO₃ --> grams KClO₃

2) 26.4g O₂ x (1 mol O₂ / 32.00 g O₂) x (2 mol KClO₃ / 3 moles O₂) x (122.65g KClO₃ / 1 mol KClO₃) = 67.46 KClO₃

I hope this helps!