Question

38 - what weight of potassium nitrate is required to 100 g of oxygen. [K-39, N-14,O=16] evolve . ​

Answer 1

Answer:

The weight of potassium nitrate required to evolve 100 grams of oxygen is 631 grams.

Explanation:

Potassium nitrate chemical formula = KNO₃

KNO₃ on heating gives KNO₂ and O₂

2KNO₃→ 2KNO₂+O₂

Here 2 moles of KNO₃ gives one mole of O₂

one mole of KNO₃ is = 101$g$

2 mole of KNO₃ is = 101×2 = 202$g$

the molar mass of O₂ = 32$g/mol$

so 32 gram of oxygen is produced from 202 gram of potassium nitrate.

⇒ one gram of oxygen is produced from = $\frac{202}{32}$

                                                                      = 6.31 gram of potassium nitrate

so to get 100-gram oxygen= 6.31×100

                                            = 631-gram potassium nitrate is needed

therefore the weight of potassium nitrate required to evolve 100 grams of oxygen is 631 grams.

Answer 2

Answer:

The weight of potassium nitrate required to evolve 100 grams of oxygen is 631 grams. so 32 gram of oxygen is produced from 202 gram of potassium nitrate. therefore the weight of potassium nitrate required to evolve 100 grams of oxygen is 631 grams.

Explanation:

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