Question

How many milliliters of oxygen gas at STP are released from heating 4.60g of calcium chlorate?
Ca(ClO3)2(s)→CaCl2(s)+3O2(g)

Answer 1

What have you done so far?
@ChiquitaM

Answer 2

Lol... everything I've done is wrong. I need help.

Answer 3

It would help if you review what you have done for the topic so far before attempting the following.
Remember, you do not want to solve one single problem. You want to solve the class of problems you have in front of you.
Here's how you would proceed:
The idea is that the mass (volume) of reactants and products are proportional to the molecular masses according to the chemical equation.
Ca(ClO3)2(s)→CaCl2(s)+3O2(g)
Note that one molar mass (in grams) of a gas measures 22.4 L in volume at STP.

Find the molar mass of \(Ca(ClO_3)_2\), m1= 40+2(35.5+3*16) (approximate;y)
molar mass of \(3O_2\), m2 = 3(16*3) (approximately)
Volume of 3 moles of oxygen = 3*22.4 L at STP
Recalculate m1 using more accurate atomic masses.
Then use the ratio:
\(\dfrac{m1}{3*22.4 L} = \dfrac{4.60}{x~Litres}\)
to solve for x in litres.
Convert to volume using 1 mole of oxygen (32 grams) occupies 22.4 L at STP.
Give answer to 3 significant figures.

Note: in the future please post in the chemistry section, and if there is no answer, post here with reference to the unanswered post. CoC requires posts to be in the appropriate section.

Answer 4

thank you soooooo very much

Answer 5

You're welcome! :)