Assuming the complete decomposition of KClO3, calculate the number of grams of O2 gas that can be attained from 46.0g of KClO3. (the products are KCl and O2. )
Please help me. I'm very confused
2006-08-25 04:43:06 · 3 answers · asked by dingdong 1 in Science & Mathematics ➔ Chemistry
The decomposition reaction of Potassium Chlorate is,
2 KClO3 --heat--> 2 KCl + 3 O2
For every 2 moles of KClO3 which decomposes, 3 moles of O2 are formed (a ratio of 3 O2 : 2 KClO3).
The molar mass of Potassium Chlorate is 122.55 g/mol.
This means that if one has 46.0 grams of KClO3, this equations to .375 moles of KClO3.
Since we already know the ratio of decomposed KClO3 to O2 formed we can use this to find the number of moles of Oxygen gas produced.
.375 moles KClO3 * (3 moles O2 / 2 moles KClO3) = .563 moles of O2.
The molar mass of O2 is 32.00 grams per mole so if one has .563 moles of O2, this equates to 18.02 grams.
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2016-12-11 15:13:45 · answer #2 · answered by rocca 4 · 0⤊ 0⤋
2KClO3-----> 2KCl + 3O2
2moles of KClO3 produces 3 moles of O2
thus 46.0 grams of KClO3 produces x grams of O2
3moles O2 x46 grams KClO3/2 moles KClO3=69 grams of O2
2006-08-25 04:53:44 · answer #3 · answered by Natasha B 4 · 0⤊ 1⤋