Question

Фосфат кальция массой 62г обработали концентрированной серной кислотой Вычислите массу образовавшейся серной кислоты

Answer 1

Ca3(PO4)2+ 3H2SO4=3CaSO4+2H3PO4 n(фосфата)=62/310=0,2моль n(h3po4)=0,4 моль m(h3po4)=0,4*98=39,2г Я так понимаю мамсу полученной фосфорной кислоты надо найти?

Answer 2

Calculation of the Mass of Sulfuric Acid Formed

To calculate the mass of sulfuric acid formed when calcium phosphate with a mass of 62g is treated with concentrated sulfuric acid, we need to consider the balanced chemical equation for the reaction.

The balanced chemical equation for the reaction between calcium phosphate (Ca3(PO4)2) and sulfuric acid (H2SO4) is as follows:

3Ca3(PO4)2 + 2H2SO4 → 6CaSO4 + 2H3PO4

From the balanced equation, we can see that 2 moles of sulfuric acid react with 3 moles of calcium phosphate to produce 2 moles of phosphoric acid (H3PO4) and 6 moles of calcium sulfate (CaSO4).

To calculate the mass of sulfuric acid formed, we need to determine the molar mass of sulfuric acid (H2SO4), which is approximately 98.09 g/mol.

Using the molar ratio from the balanced equation, we can set up the following proportion:

(2 moles H2SO4 / 3 moles Ca3(PO4)2) = (x g H2SO4 / 62 g Ca3(PO4)2)

Solving for x, the mass of sulfuric acid formed, we get:

x = (2 moles H2SO4 / 3 moles Ca3(PO4)2) * (62 g Ca3(PO4)2)

x ≈ 41.33 g

Therefore, the mass of sulfuric acid formed when 62 g of calcium phosphate is treated with concentrated sulfuric acid is approximately 41.33 g.

Please note that the above calculation assumes that the reaction goes to completion and that there are no other side reactions or losses during the process.