Question

Вычислите массу азотной кислоты, которую можно получить с натрий нитрата массой 17 г в результате его взаимодействия с концентрированной серной кислотой, если массовая доля выхода кислоты составляет 0,96. ПЛИИИИЗ,РЕШИТЕ

Answer 1

17г                                                  x

NaNO3 + H2SO4(конц) = HNO3 + NaHSO4

85г                                                  63г

m=M*n

m(NaNO3)по р-ции=85г

m(HNO3)по р-ции=63г

m(NaNO3)по усл=17г

 

x=17*63\85 = 12.6г - m(HNO3)

 

Ответ: 12.6 г

 

 

Answer 2

To calculate the mass of nitric acid that can be obtained from the reaction between sodium nitrate and concentrated sulfuric acid, we need to consider the balanced chemical equation for the reaction. The balanced equation for the reaction is as follows:

2 NaNO3 + H2SO4 -> 2 HNO3 + Na2SO4

From the balanced equation, we can see that 2 moles of sodium nitrate (NaNO3) react with 1 mole of sulfuric acid (H2SO4) to produce 2 moles of nitric acid (HNO3) and 1 mole of sodium sulfate (Na2SO4).

To find the mass of nitric acid produced, we need to determine the number of moles of sodium nitrate used. We can do this by dividing the given mass of sodium nitrate (17 g) by its molar mass.

The molar mass of sodium nitrate (NaNO3) can be calculated as follows: - The atomic mass of sodium (Na) is 22.99 g/mol. - The atomic mass of nitrogen (N) is 14.01 g/mol. - The atomic mass of oxygen (O) is 16.00 g/mol. - Since there are three oxygen atoms in sodium nitrate, we multiply the atomic mass of oxygen by 3. - Adding up the atomic masses, we get: 22.99 + 14.01 + (16.00 * 3) = 85.00 g/mol.

Now, we can calculate the number of moles of sodium nitrate: Number of moles = Mass / Molar mass = 17 g / 85.00 g/mol = 0.20 mol.

Since the molar ratio between sodium nitrate and nitric acid is 2:2, the number of moles of nitric acid produced is also 0.20 mol.

Next, we need to calculate the mass of nitric acid