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: 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

Hiii.




 
Answer: 57.55 g

Explanation

1. Write the balanced equation

\mathbb{2KCLO_3 \rightarrow 2KCL + O_2}2KCLO3​→2KCL+O2​ 

2. Calculate the atomic mass

\begin{lgathered}\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\end{lgathered}2[39+35.5+(16×2)→2(39+35.5)+3(32)2(39+35.5+48)→2×74.5+962×122.5→149+96245→245​

3. From the equation we can find that to produce 149 g of KCL we need 245 g of \mathbb{KCLO_3}KCLO3​ . 

Hence, 149 g of \mathbb{KCL}KCL is produced by \large\bf{245}245 g of \mathbb{KCLO_3}KCLO3​ . 

1 g of \mathbb{KCL}KCL is produced by \large \bf \frac{245}{149}149245​ g of \mathbb{KCLO_3}KCLO3​ .

35 g of \mathbb{KCL}KCL is produced by \large \bf \frac{245}{149} \times 35 = 57.55149245​×35=57.55 g of \mathbb{KCLO_3}KCLO3​ .




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