Question

Составьте уравнения химических реакций, соответствующие данной схеме: P->P2O5->H3PO4. Рассчитайте, какая масса фосфора потребуется для получения 2 моль фосфорной кислоты. Заранее спасибо!

Answer 1

4Р+5О2 --> 2Р2О5

Р2О5+3Н2О --> 2Н3РО4

для получения 2 моль кислоты требуется 1 моль оксида фосфора (V)

по 1 уравнению для получения 1 моль оксида требуется 2 моль фосфора (расход 1 к 2) ---> надо 2 моль фосфора

m(P)=n*M=2моль*31г/моль=62г

Ответ 62г

Answer 2

Chemical Reactions:

The given scheme represents the chemical reactions involved in the conversion of phosphorus (P) to phosphoric acid (H3PO4). The reactions can be written as follows:

1. Conversion of phosphorus to phosphorus pentoxide (P2O5): ``` 4P + 5O2 -> 2P2O5 ```

2. Conversion of phosphorus pentoxide to phosphoric acid: ``` P2O5 + 3H2O -> 2H3PO4 ```

Calculation of Phosphorus Mass:

To calculate the mass of phosphorus required to obtain 2 moles of phosphoric acid (H3PO4), we need to use the stoichiometry of the balanced chemical equation.

From the balanced equation, we can see that 2 moles of phosphoric acid (H3PO4) are produced from 4 moles of phosphorus (P). Therefore, the molar ratio between phosphorus and phosphoric acid is 4:2, or simply 2:1.

Since the molar mass of phosphorus is approximately 31 grams per mole, we can calculate the mass of phosphorus required as follows:

Mass of phosphorus = (2 moles of H3PO4) * (4 moles of P / 2 moles of H3PO4) * (31 grams per mole of P)

Simplifying the equation:

Mass of phosphorus = 2 * 4 * 31 grams
Mass of phosphorus = 248 grams

Therefore, to obtain 2 moles of phosphoric acid, approximately 248 grams of phosphorus are required.

Please note that the molar mass of phosphorus and the stoichiometry of the reaction are approximate values and may vary slightly depending on the specific isotopes and reaction conditions.