**Lowell High
School Chemistry**

**Stoichiometry
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**Practice
Problems Here:**

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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