Math, asked by Raflem, 1 year ago

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

Answers

Answered by Anonymous
2
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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}.
Answered by BrainlyHeart751
2

Answer:


Step-by-step explanation:

Assuming complete thermal decomposition of KClO3 i.e. potassium chlorate the following reaction takes place…


2KClO3 → 2KCl + 3O2


Hence it's clearly visible that the molar ration between KClO3 and O2 is 2:3. Intuitively speaking u need 2 mol of compound to react with 3 mol of O2.


Now the molar volume at STP for an ideal gas is 22.4 dm3/mol (assuming O behaves ideally we shall take this number). Intuitively speaking one mol of ideal gas occupies 22.4 dm3 volume. Hence if u have 6.72 dm3 of ideal gas u have 0.3 mol of that gas (6.72/22.4).


Since we have the molar ratio as 2:3 and we have no of moles of O2 as 0.3, the no of moles of KClO3 can be calculated and it turns out to be 0.2 mol.


Mass = No. Of moles * Molar mass ofor substance


Molar mass of KClO3 is roughly 122.5 g/mol.


Hence the mass required will be = 0.2 mol *122.5 g/mol


And u get the mass as 24.5 g


Hope this works…

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