Question

Сколько граммов ортофосфорной кислоты H3PO4 образуется при взаимодействии 7,2 г оксида фосфора (V) с водой?

Answer 1

                                                        P2O5+3H2O=2H3PO4

                            масса                   7,2                       Х

            моль                                       1                        2

        моляр.масса                        142                     98

         кол-во в-ва                            142                    196

 

Х=7,2*196/142=9,9гр

Answer 2

Calculating the Amount of Orthophosphoric Acid (H3PO4) Formed

To calculate the amount of orthophosphoric acid (H3PO4) formed when 7.2 g of phosphorus(V) oxide reacts with water, we can use the balanced chemical equation for the reaction:

P4O10 + 6H2O → 4H3PO4

From the balanced equation, we can see that 1 mole of phosphorus(V) oxide (P4O10) reacts with 6 moles of water (H2O) to produce 4 moles of orthophosphoric acid (H3PO4).

First, we need to calculate the number of moles of phosphorus(V) oxide using its molar mass.

The molar mass of P4O10 is 283.89 g/mol. The molar mass of H3PO4 is 97.99 g/mol.

Using this information, we can proceed with the calculation.

Calculation Steps:

1. Calculate the number of moles of P4O10: - Number of moles = Mass / Molar mass - Number of moles = 7.2 g / 283.89 g/mol

2. Use the mole ratio from the balanced chemical equation to find the moles of H3PO4 produced: - Moles of H3PO4 = (Moles of P4O10) * (4 moles of H3PO4 / 1 mole of P4O10)

3. Convert the moles of H3PO4 to grams: - Mass of H3PO4 = Moles of H3PO4 * Molar mass of H3PO4

Let's calculate the amount of orthophosphoric acid (H3PO4) formed.