**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 KClO_{3} is heated and decomposes completely, what mass of oxygen gas is produced?

KClO_{3} (s) –> KCl (s) + O_{2} (g) [unbalanced]

**Answer to Practice Problem #1:**

**1.** **Balance your equation first.**

**2**KClO_{3} (s) –> **2**KCl (s) + **3**O_{2} (g) [balanced]

**2.** **Convert from grams to moles using molar mass.**

To get moles from grams of potassium chlorate (KClO_{3}): [molar mass = 122.55 g/mol]

138.6 g KClO_{3} X ( 1 mol / 122.55 g ) = 1.131 mol KClO_{3}

**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 KClO_{3} is the only reactant, it is the limiting reagent. You use the moles of KClO_{3} 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 KClO_{3} X ( 3 mol O_{2} / 2 mol KClO_{3} ) = 1.696 mol O_{2} 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 (O_{2}): [molar mass = 32 g/mol]

1.696 mol O_{2 } X ( 32 g / 1 mol ) = **54.286 g O _{2} produced**

**Answer to Problem:** **54.286 g O _{2} produced**

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