Question

Сколько граммов азотной кислоты образуется при взаимодействии 170 граммов нитрата натрия с избытком серной кислоты.Помогите пожалуйста!

Answer 1

       170 г                                                                             Х г
     2NaNO3         +         H2SO4     ->     Na2SO4     +     2HNO3
  n=2 моль                                                                     n=2 моль
М = 85 г/моль                                                               М = 63 г/моль
m=170 г                                                                         m=126 г

170 г NaNO3    -    Х г HNO3
170 г NaNO3    -   126 г HNO3
m(HNO3) = 170 * 126 / 170 = 126 г

Answer 2

Calculating the Amount of Nitric Acid Formed

To calculate the amount of nitric acid formed when 170 grams of sodium nitrate reacts with an excess of sulfuric acid, we can use the balanced chemical equation for the reaction between sodium nitrate and sulfuric acid.

The balanced chemical equation for the reaction is: ``` 2 NaNO3 + H2SO4 → 2 HNO3 + Na2SO4 ```

From the balanced chemical equation, we can see that 2 moles of sodium nitrate react with 1 mole of sulfuric acid to produce 2 moles of nitric acid. Now, let's calculate the amount of nitric acid formed.

Calculation

1. Calculate the moles of sodium nitrate: - Moles of NaNO3 = Mass of NaNO3 / Molar mass of NaNO3 - Moles of NaNO3 = 170 g / (22.99 g/mol + 14.01 g/mol + 3(16.00 g/mol)) - Moles of NaNO3 = 170 g / 85 g/mol - Moles of NaNO3 = 2 moles

2. Calculate the moles of nitric acid formed: - Moles of HNO3 = Moles of NaNO3 * (2 moles of HNO3 / 2 moles of NaNO3) - Moles of HNO3 = 2 moles * (2 moles / 2 moles) - Moles of HNO3 = 2 moles

3. Calculate the mass of nitric acid formed: - Mass of HNO3 = Moles of HNO3 * Molar mass of HNO3 - Mass of HNO3 = 2 moles * (1(1.01 g/mol) + 14.01 g/mol + 3(16.00 g/mol)) - Mass of HNO3 = 2 moles * 63.01 g/mol - Mass of HNO3 = 126.02 grams

The amount of nitric acid formed when 170 grams of sodium nitrate reacts with an excess of sulfuric acid is 126.02 grams.

[[4 #]]