Lowell High School Chemistry
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Practice Problem #1:
Oxygen gas can be produced by decomposing potassium chlorate using the reaction below. If 138.6 g of KClO3 is heated and decomposes completely, what mass of oxygen gas is produced?
KClO3 (s) –> KCl (s) + O2 (g) [unbalanced]
Answer to Practice Problem #1:
1. Balance your equation first.
2KClO3 (s) –> 2KCl (s) + 3O2 (g) [balanced]
2. Convert from grams to moles using molar mass.
To get moles from grams of potassium chlorate (KClO3): [molar mass = 122.55 g/mol]
138.6 g KClO3 X ( 1 mol / 122.55 g ) = 1.131 mol KClO3
3. Determine the limiting reagent [if necessary] (Use mole ratios to figure out.)
* Not necessary to determine because there is only 1 reactant in this decomposition reaction.
4. Use ratios to find the moles of the reactant or product you need to find.
Since KClO3 is the only reactant, it is the limiting reagent. You use the moles of KClO3 if you have to calculate how many moles of oxygen gas (the product you’re trying to find) are being produced. Use your mole ratios from your balanced equation:
1.131 mol KClO3 X ( 3 mol O2 / 2 mol KClO3 ) = 1.696 mol O2 produced
5. Convert from moles back to grams of the new substance using that substance’s molar mass.
To get grams from moles of carbon (O2): [molar mass = 32 g/mol]
1.696 mol O2 X ( 32 g / 1 mol ) = 54.286 g O2 produced
Answer to Problem: 54.286 g O2 produced