Question

К раствору содержащему 12,6 г азотной кислоты добавили раствор содержащий 7,2 гидроксида натрия.Сколько гидроксида по массе потребуется для полной нейтрализации?

Answer 1

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Answer 2

Calculating the Amount of Sodium Hydroxide Needed for Neutralization

To calculate the amount of sodium hydroxide (NaOH) needed for the complete neutralization of nitric acid (HNO3), we can use the balanced chemical equation for the reaction between nitric acid and sodium hydroxide:

HNO3 + NaOH → NaNO3 + H2O

From the equation, we can see that 1 mole of nitric acid reacts with 1 mole of sodium hydroxide. We can use the molar masses of the substances to convert the given masses of nitric acid and sodium hydroxide into moles, and then determine the stoichiometric ratio to find the amount of sodium hydroxide needed.

Calculation Steps

1. Calculate the moles of nitric acid (HNO3) and sodium hydroxide (NaOH) using their respective molar masses. 2. Determine the stoichiometric ratio of nitric acid to sodium hydroxide. 3. Use the stoichiometric ratio to calculate the amount of sodium hydroxide needed for complete neutralization.

Calculation

1. Moles of nitric acid (HNO3): - Given mass of nitric acid = 12.6 g - Molar mass of nitric acid (HNO3) = 63.01 g/mol - Moles of HNO3 = Given mass / Molar mass = 12.6 g / 63.01 g/mol = 0.2 moles

2. Moles of sodium hydroxide (NaOH): - Given mass of sodium hydroxide = 7.2 g - Molar mass of sodium hydroxide (NaOH) = 40.00 g/mol - Moles of NaOH = Given mass / Molar mass = 7.2 g / 40.00 g/mol = 0.18 moles

3. Stoichiometric ratio: - From the balanced chemical equation, 1 mole of HNO3 reacts with 1 mole of NaOH.

4. Amount of sodium hydroxide needed for complete neutralization: - Since the stoichiometric ratio is 1:1, the amount of sodium hydroxide needed is the same as the moles of nitric acid, which is 0.2 moles.

Therefore, 0.2 moles of sodium hydroxide are needed for the complete neutralization of the given amount of nitric acid.