Chemistry, 12.08.2020 08:01 gyexisromero10

How many moles of potassium in 73.56g of potassium chlorate (V) (KClO 3 )?

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Answer from: samirahscott

73.56 g of potassium chlorate (v), KClO₃ contains 0.6 mole of potassium, K.

To know the exact number of mole of potassium (K) in 73.56 g of potassium chlorate (v), KClO₃ do the following:

Step 1:

Determination of the number of mole in 73.56 g of potassium chlorate (v), KClO₃

Mass of KClO₃ = 73.56 g

Molar mass of KClO₃ = 39 + 35.5 + (16×3)

= 39 + 35.5 + 48

= 122.5 g/mol

Mole of KClO₃ =?

$Mole =\frac{mass}{molar mass}\\\\Mol KClO_{3} = \frac{73.56}{122.5}$

Mole of KClO₃ = 0.6 moleStep 2:

Determination of the number of mole of potassium, K in 0.6 mole (i.e 73.56 g) of KClO₃

Considering the molecular formula of potassium chlorate (v), KClO₃, we can see that:

1 mole of KClO₃ contains 1 mole of K.

Therefore, 0.6 mole of KClO₃ will also contain 0.6 mole of K.

Therefore, we can conclude that 73.56 g of KClO₃ contains 0.6 mole of potassium, K.

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Answer from: axell13965

Approximately $0.6003\; \rm mol$ formula units.

Explanation:

Formula Mass of KClO₃

Look up the relative atomic mass data for $\rm K$ , $\rm Cl$ , and $\rm O$ on a modern periodic table:

$\rm K$ :

. $\rm Cl$ : 35.45

. $\rm O$ : 15.999

The relative atomic mass of an element is numerically equal to the mass (in grams) of one mole of its atoms. For example, the relative atomic mass of $\rm K$ is 39.908 . Therefore, the mass of one mole of $\rm K\!$ atoms should be $39.908\; \rm g$ .

Each $\rm KClO_3$ "formula" unit includes one $\rm K$ atom, one $\rm Cl$ atom, and three $\rm O$ atoms. Therefore, one mole of $\rm KClO_3\!$ formula units would include:

one mole of $\rm K$ atoms, one mole of $\rm Cl$ atoms, and three moles of $\rm O$ atoms.

From the relative atomic mass of these three elements:

The mass of one mole of $\rm K$ atoms would be $39.908\; \rm g$ .The mass of one mole of $\rm Cl$ atoms would be $35.45\; \rm g$ .The mass of three moles of $\rm O$ atoms would be $3 \times 15.999\; \rm g = 47.997\; \rm g$ .

Combining these three parts should give the mass of one mole of $\rm KClO_3$ formula units:

$\begin{aligned}& M(\mathrm{KClO_3}) \\ &= 39.908 + 35.45 + 3 \times 15.999 \\ &= 122.545\; \rm g \cdot mol^{-1}\end{aligned}$ .

Number of moles of KClO₃ formula units in the sample

The formula mass of $\rm KClO_3$ is $M(\mathrm{KClO_3}) = 122.545\; \rm g \cdot mol^{-1}$ , meaning that the mass of one mole of $\rm KClO_3\!$ formula units would be $122.545\; \rm g\!$ .

The mass of this $\rm KClO_3\!\!$ sample is $m(\mathrm{KClO_3}) = 73.56\; \rm g$ . The number of moles of formula $\rm KClO_3\!$ units in this sample would be:

$\begin{aligned}n(\mathrm{KClO_3}) &= \frac{m(\mathrm{KClO_3})}{M(\mathrm{KClO_3})} \\ &= \frac{73.56\; \rm g}{122.545\; \rm g \cdot mol^{-1}} \approx 0.6003\; \rm mol\end{aligned}$ .

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