Question

Каждую массу ортофосфата натрия можно получить при взаимодействии 200 г ортофосфарной кислоты с 250 г гидроксида натрия? помогите пожалуйста

Answer 1

Дано:
m(H3PO4) = 200 g
m(NaOH) = 250 g
m(Na3PO4) - ?

n(H3PO4) = 200 g/98 g/mol
n(H3PO4) = 2.04 mol
n(NaOH) = 250 g/40g/mol =  6.25 mol
2,04:6,25 = 1:3,06

H3PO4 + 3NaOH = Na3PO4 + 3H2O

Из уравнения реакции и произведенным расчетам следует, что гидроксид натрия взят в избытке и n(Na3PO4) = n(H3PO4)=2.04 моль
M(Na3PO4) = 164 г/моль
m(Na3PO4) = 2.04 моль*164 г/моль = 334,56 г
Ответ: 334,56 г

Answer 2

  200 g     250g         x mol
H3PO4+3NaOH = Na3PO4+3H2O
 1 mol      3 mol        1 mol

200 g / 250 g ≠ 1 mol / 3 mol, отсюда делаем вывод, что гидроксид натрия прореагировал не полностью, поэтому в дальнейших расчётах будем использовать ортофосфатную кислоту.

n(H3PO4) = m(H3PO4) / M(H3PO4) = 200 g / 98 g/mol = 2.04 mol

2.04 mol / 1 mol = x mol / 1 mol
x = (2.04 mol * 1 mol) / 1 mol = 2.04 mol, n(Na3PO4) = 2.04 mol

m(Na3PO4) = n(Na3PO4) * M(Na3PO4) = 2.04 mol * 164 g/mol = 334.56 g

m(Na3PO4) = 334.56 g

Answer 3

Calculation of Sodium Orthophosphate Mass

To calculate the mass of sodium orthophosphate that can be obtained from the reaction between 200 g of orthophosphoric acid and 250 g of sodium hydroxide, we need to determine the stoichiometry of the reaction.

The balanced chemical equation for the reaction between orthophosphoric acid (H₃PO₄) and sodium hydroxide (NaOH) is as follows:

2 H₃PO₄ + 3 NaOH → Na₃PO₄ + 3 H₂O

From the balanced equation, we can see that 2 moles of orthophosphoric acid react with 3 moles of sodium hydroxide to produce 1 mole of sodium orthophosphate.

To calculate the moles of orthophosphoric acid and sodium hydroxide, we can use their molar masses. The molar mass of orthophosphoric acid (H₃PO₄) is approximately 98 g/mol, and the molar mass of sodium hydroxide (NaOH) is approximately 40 g/mol.

Using these molar masses, we can calculate the moles of orthophosphoric acid and sodium hydroxide as follows:

Moles of orthophosphoric acid = mass of orthophosphoric acid / molar mass of orthophosphoric acid Moles of sodium hydroxide = mass of sodium hydroxide / molar mass of sodium hydroxide

Let's calculate the moles of orthophosphoric acid and sodium hydroxide:

Moles of orthophosphoric acid = 200 g / 98 g/mol = 2.04 molMoles of sodium hydroxide = 250 g / 40 g/mol = 6.25 mol According to the stoichiometry of the balanced equation, 2 moles of orthophosphoric acid react with 3 moles of sodium hydroxide to produce 1 mole of sodium orthophosphate.

Since the reaction is limited by the reactant that is present in the smallest amount, we can determine the limiting reactant by comparing the mole ratios of orthophosphoric acid and sodium hydroxide.

The mole ratio of orthophosphoric acid to sodium hydroxide is 2:3. Therefore, for every 2.04 moles of orthophosphoric acid, we need 3/2.04 = 1.47 moles of sodium hydroxide to react completely.

Since we have 6.25 moles of sodium hydroxide, which is greater than the required 1.47 moles, sodium hydroxide is in excess. This means that orthophosphoric acid is the limiting reactant.

From the stoichiometry of the balanced equation, we know that 2 moles of orthophosphoric acid react to produce 1 mole of sodium orthophosphate.

Therefore, the moles of sodium orthophosphate that can be obtained from 2.04 moles of orthophosphoric acid are:

Moles of sodium orthophosphate = 2.04 mol / 2 = 1.02 mol

To calculate the mass of sodium orthophosphate, we can use its molar mass. The molar mass of sodium orthophosphate (Na₃PO₄) is approximately 164 g/mol.

Mass of sodium orthophosphate = moles of sodium orthophosphate * molar mass of sodium orthophosphate Mass of sodium orthophosphate = 1.02 mol * 164 g/mol = 167.28 g

Therefore, the mass of sodium orthophosphate that can be obtained from the reaction between 200 g of orthophosphoric acid and 250 g of sodium hydroxide is approximately 167.28 g.

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

I hope this helps! Let me know if you have any further questions.