Question

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

Answer 1

$<b>$
Answer: 57.55 g

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

$\huge \mathfrak {Answer:-}$

$\bold{Reaction:-}$

$2KClO_{3}$ => $2KCl$ + $O_{2}$

$\bold{Gram\:mass\:of:}$

$2KClO_{3}$

K = 39, Cl = 35.5, O = 16

$\bold{So,}$

$= 2(39 + 35.5 + 48)$

$= 2 \times 122.5$

$= 245$

$\bold{This\:means\:that:-}$

${245\:g}$ of potassium chlorate was used.

$\bold{Now,}$

Mass of potassium chloride is:

$= 2(39 + 35.5)$

$= 2 \times 74.5$

${=\:149\:g}$

$\bold{Hence,}$

${149\:g}$ of KCl is produced per ${245\:g}$.

$\bold{And,}$

${1\:g}$ is produced per $\frac{245}{149} g$

$\bold{So,}$

${35\:g}$ is produced per:

$= \frac{245}{149} \times 35$

${=\:57.55\:g}$

$\bold{Therefore,}$

The mass of potassium chlorate required to produce ${35\:g}$ of potassium chloride on complete decomposition is ${57.55\:g.}$