Calculating Oxygen Gas Released from Heating Calcium Chlorate

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Heating 5.00 grams of calcium chlorate releases approximately 1620 milliliters of oxygen gas. The reaction involves the decomposition of calcium chlorate into calcium chloride and oxygen, with a mole ratio of 3 moles of O2 for every mole of Ca(ClO3)2. The calculations require converting grams of calcium chlorate to moles, applying the mole ratio, and then converting moles of oxygen to volume using the ideal gas law. The initial calculation was incorrect due to a misunderstanding of the mole conversion process. The correct approach yields a final volume of about 1617 to 1620 mL of oxygen gas.
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Homework Statement


How many millilitres of oxygen gas are released from heating 5.00 grams of calcium chlorate?

Homework Equations



Ca(ClO3)2→CaCl2+3O2

Ca(ClO3)2 =207.7g
CaCl2=110.98g
O2=22.4 g/L
Mole ratio= 3mol O2/1mol Ca(Clo3)2

The Attempt at a Solution


This is what I have but it is not coming out right and I just can't figure out where I am going wrong. I am having trouble with these kinds of problems.

(207.7 g/ 1 mol Ca(Clo3)2) X (1mol Ca(Clo3)2/3 mol O2) X (22.4 L O2/ 1 mol O2)= 1550.8
(this can't be right the answer is supposed to be 1620)
 
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Moles of Calcium Chlorate (Given Mass / Molar Mass)* Moles of dioxygen per mole of Calcium Chlorate (3) * Volume of 1 mole dioxygen (22.4 L / mol).

The answer is approximately 1617. (1620 if you approximately calculate it manually)
 
I now see where I went wrong! thank you :)
 
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