Question

calculate the mass of potassium chlorate required to produce 35 g of potassium chloride on complete decomposition

Answer 1

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Answer:

Explanation

1. Write the balanced equation

$\mathbb{2KCLO_3 \rightarrow 2KCL + O_2}$

2. Calculate the atomic mass

$\rm {2[ 39 + 35.5 + (16 \times 2)] \rightarrow 2(39 + 35.5) + 3(32)} \\ \rm2(39 + 35.5 + 48) \rightarrow 2 \times 74.5 + 96 \\ \rm 2 \: \times \: 122.5 \rightarrow 149 + 96 \\ \rm245 \rightarrow 245$

3. From the equation we can find that to produce 149 g of KCL we need 245 g of $\mathbb{KCLO_3}$.

Hence, 149 g of $\mathbb{KCL}$ is produced by $\large\bf{245}$ g of $\mathbb{KCLO_3}$.

1 g of $\mathbb{KCL}$ is produced by $\large \bf \frac{245}{149}$ g of $\mathbb{KCLO_3}$.

35 g of $\mathbb{KCL}$ is produced by $\large \bf \frac{245}{149} \times 35 = 57.55$ g of $\mathbb{KCLO_3}$.

Answer 2

GIVE ME THANKS AND BRAINLIEST ANSWER
2KClO3 -> 2KCl + 3O2

35g of KClO3 (Mol. Mass 122.55g/mole) = (35/122.55) =0.2857 moles

Each mole of KClO3 decomposes to produce 1 mole of KCl (Mol. Mass = 74.55g/mole)

Then moles of KCl produced =0.2857  moles

Mass of this amount of KCl =(0.2857  x74.55) = 21.29g approx.

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