How many grams of potassium chlorate decompose to

Chapter 6, Problem 44P

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QUESTION:

How many grams of potassium chlorate decompose to potassium chloride and \(638 mL\) of \(\mathrm{O}_{2}\) at \(128^{\circ} \mathrm{C}\) and \(752 torr\)?

\(2 \mathrm{KCIO}_{3}(s) \rightarrow 2 \mathrm{KCI}(s)+3 \mathrm{O}_{2}(g)\)

Equation Transcription:

O2

128°C

Text Transcription:

638 mL

O_2

128°C

752 torr

2KCIO_3(s) rightarrow 2KCI(s)+3O_2(g)

Questions & Answers

QUESTION:

How many grams of potassium chlorate decompose to potassium chloride and \(638 mL\) of \(\mathrm{O}_{2}\) at \(128^{\circ} \mathrm{C}\) and \(752 torr\)?

\(2 \mathrm{KCIO}_{3}(s) \rightarrow 2 \mathrm{KCI}(s)+3 \mathrm{O}_{2}(g)\)

Equation Transcription:

O2

128°C

Text Transcription:

638 mL

O_2

128°C

752 torr

2KCIO_3(s) rightarrow 2KCI(s)+3O_2(g)

ANSWER:

Solution 44P

Here we have to find out the mass of  potassium chlorate in grams.

Step 1

Given:

The balanced chemical equation for decomposition of potassium chlorate is given below,

2 KClO3 (s) → 2KCl (s) + 3O2 (g)

Volume of O2 = 638 mL of O2 at  = 638 mL = 0.638 L

Temperature = 128°C = 273 + 128 = 401 K

Pressure =  752 torr = 752 torr = 0.989 atm

n =

Where P = pressure

        V = Volume

        T = temperature

        n = number of moes

        R = gas constant i.e 0.0821 L atm mol-1 k-1

Now substituting the values, the number of moles  can be calculated as,

n =

   = = 0.019 mol O2

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