Question

4 г кальцію вступило у реакцію із ортофосфатною кислотою. При цьому утворились сіль і водень. Визначити масу водню, що при цьому утворився.

Answer 1

Ответ:

Газ в m не измеряется, измеряется в объёме V

Ответ: V(H2) = 2,24 л

Объяснение:

Дано:

m( Ca)=4г

V(H2) -?

Решение:

3Ca+2H3PO4-->Ca3(PO4)2+3H2 (газ)

1) n= m/Mr

n(Ca)= 4/40=0,1 моль

по условию n(Ca)= 0,1 моль n(H2)= X моль

по уравнению n(Ca)=3 моль n(H2)=3 моль

2) Составляем пропорцию:

n(H2)= $\frac{0.1 \times 3}{3} = 0.1 \: моль$

3) V = n×Vm

V(H2)= 0.1×22.4=2,24 л

Ответ: V(H2) = 2,24 л

Answer 2

Reaction of Calcium with Orthophosphoric Acid

When 4 grams of calcium reacts with orthophosphoric acid, a salt and hydrogen are formed. To determine the mass of hydrogen produced in this reaction, we can use the stoichiometry of the reaction and the molar mass of calcium.

The balanced chemical equation for the reaction between calcium and orthophosphoric acid is:

3Ca + 2H3PO4 → Ca3(PO4)2 + 3H2

Calculation of the Mass of Hydrogen Produced

To calculate the mass of hydrogen produced, we can use the stoichiometry of the reaction. From the balanced chemical equation, we can see that 3 moles of hydrogen are produced for every 2 moles of orthophosphoric acid reacted.

Given that the molar mass of orthophosphoric acid (H3PO4) is approximately 98 g/mol, we can calculate the moles of orthophosphoric acid present in 4 grams:

moles of H3PO4 = mass / molar mass = 4 g / 98 g/mol

Using the stoichiometry of the reaction, we can then calculate the moles of hydrogen produced:

moles of H2 = (3/2) * moles of H3PO4

Finally, we can determine the mass of hydrogen produced using the calculated moles of hydrogen and the molar mass of hydrogen (approximately 2 g/mol).

Conclusion

By following these calculations, we can determine the mass of hydrogen produced when 4 grams of calcium reacts with orthophosphoric acid.