Question

К 400г грамм раствора ортофосфорной кислоты с массовой долей H3PO4 24,5% добавили оксид фосфора, полученный сжиганием 62г фосфора, и раствор прокипятили. Вычислите объём гидроксида натрия с массовой долей NaOH 5% (пл. 1.05 г/мл), который потребуется для превращения всей содержавшийся в растворе кислоты в гидрофосфат натрия.

Answer 1

1. 8H3PO4+P4O10=.----80-100t =6H4P2O7
2. 4P+5O2=2P2O5

3. 2Na+H4P2O7=Na2H2P2O7+2H2O

Mr(P)=31г/моль
n(P)=m/Mr=62/31=2моль
Составляемый пропорцию по первому уравнению:
2/4=х/2. Х=2*2/4=1моль(P2O5)

m(H3PO4)=400*0,245=98гр
Mr(H3PO4)=98г/моль
n(H3PO4)= m/Mr=98/98=1моль
Составляемый пропорцию по. 1-му уравнению:
1/8=х/6; 1*6/8=0,75моль H4P2O7

Mr(H4P2O7)=178г/моль
m(H4P2O7)=178*0,75=133,5гр
По третьему уравнению реакции мы видим,что
На 2моль NaOH идет 1 моль кислоты:
n(NaOH)=0,75*2=1,5моль
Mr(NaOH)=40г/моль
m(NaOH)=n*Mr=1,5*40=60грамм
m(раствора)(NaOH)=60/0,05=1200гр
V(р-ра)(NaOH)=m/p=1200/1,05=1143мл

Answer 2

Task: Calculate the volume of sodium hydroxide solution needed to convert all the orthophosphoric acid in the solution to sodium dihydrogen phosphate.

To calculate the volume of sodium hydroxide solution needed to convert all the orthophosphoric acid in the solution to sodium dihydrogen phosphate, we need to use the balanced chemical equation for the reaction between orthophosphoric acid (H3PO4) and sodium hydroxide (NaOH):

H3PO4 + NaOH -> NaH2PO4 + H2O

From the equation, we can see that one mole of orthophosphoric acid reacts with one mole of sodium hydroxide to produce one mole of sodium dihydrogen phosphate and one mole of water.

To find the number of moles of orthophosphoric acid in the solution, we can use the mass and molar mass of orthophosphoric acid:

Given: - Mass of orthophosphoric acid solution = 400 g - Mass fraction of H3PO4 in the solution = 24.5%

First, we need to calculate the mass of orthophosphoric acid in the solution:

Mass of H3PO4 = Mass of solution * Mass fraction of H3PO4
             = 400 g * 0.245
             = 98 g

Next, we need to calculate the number of moles of orthophosphoric acid:

Molar mass of H3PO4 = 3 * (1.00784 + 15.999) + 1.00784 * 4
                  = 97.994 g/mol

Number of moles of H3PO4 = Mass of H3PO4 / Molar mass of H3PO4
                        = 98 g / 97.994 g/mol
                        ≈ 1 mol

Since the stoichiometry of the reaction is 1:1 between orthophosphoric acid and sodium hydroxide, we know that we need 1 mole of sodium hydroxide to react with 1 mole of orthophosphoric acid.

To calculate the volume of sodium hydroxide solution needed, we can use the molar concentration of sodium hydroxide and the number of moles of sodium hydroxide:

Given: - Mass fraction of NaOH in the solution = 5% - Density of sodium hydroxide solution = 1.05 g/mL

First, we need to calculate the mass of sodium hydroxide in the solution:

Mass of NaOH = Mass of solution * Mass fraction of NaOH
             = 400 g * 0.05
             = 20 g

Next, we need to calculate the number of moles of sodium hydroxide:

Molar mass of NaOH = 22.98977 + 15.999 + 1.00784
                  = 39.997 g/mol

Number of moles of NaOH = Mass of NaOH / Molar mass of NaOH
                        = 20 g / 39.997 g/mol
                        ≈ 0.5 mol

Since the stoichiometry of the reaction is 1:1 between sodium hydroxide and orthophosphoric acid, we know that we need 0.5 moles of sodium hydroxide to react with 1 mole of orthophosphoric acid.

Finally, we can calculate the volume of sodium hydroxide solution needed:

Volume of NaOH solution = (Number of moles of NaOH / Molar concentration of NaOH) / Density of NaOH solution
                       = (0.5 mol / 0.05 mol/L) / 1.05 g/mL
                       = 10 mL

Therefore, 10 mL of sodium hydroxide solution with a mass fraction of NaOH of 5% (density of 1.05 g/mL) is needed to convert all the orthophosphoric acid in the solution to sodium dihydrogen phosphate.

Please note that the calculations provided are based on the given information and assumptions.