The decomposition of potassium chlorate yields oxygen gas. if the yield is 85%, how many grams of kclo3 are needed to produce 15.0 l of o2 ?

2kclo3(s) -- > 2kcl(s) + 3 o2(g)

Explanation:

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At first, the undergoing chemical reaction is:

As 15.0 L of oxygen were produced, one computes this amount in moles considering STP conditions (0°C and 1 atm) for the ideal gas equation as shown below:

Now, since those moles are the actual obtained moles, one computes the theoretical moles of oxygen which are more precisely associated with the needed grams of potassium chlorate as follows:

Finally, we apply the stoichiometry to determine the moles of potassium chlorate as shown below:

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