Question

Вычислить массу Nano3 Полученных при реакции 20 грамм NAOH и 126 грамм HNO3

Answer 1

NaOH+HNO3=NaNO3+H2O

n(NaOH)=20г/40г/моль=0,5моль (н)
n(HNO3)=126г/63г/моль=2моль (и)

Решаем задачу по недостатку:
n(NaNO3)=n(HNO3)
m(NaNO3)=0.5моль*85г/моль=42,5

Answer 2

Calculating the Mass of Nano3 Produced

To calculate the mass of NaNO3 produced in the reaction between 20 grams of NaOH and 126 grams of HNO3, we can use the stoichiometry of the reaction and the given masses of the reactants.

The balanced chemical equation for the reaction is: 2 NaOH + HNO3 → NaNO3 + H2O

Using the molar masses of the compounds: - Molar mass of NaOH (Na = 22.99 g/mol, O = 16.00 g/mol, H = 1.01 g/mol): 22.99 + 16.00 + 1.01 = 40.00 g/mol - Molar mass of HNO3 (H = 1.01 g/mol, N = 14.01 g/mol, O = 16.00 g/mol): 1.01 + 14.01 + (3 * 16.00) = 63.02 g/mol - Molar mass of NaNO3 (Na = 22.99 g/mol, N = 14.01 g/mol, O = 16.00 g/mol): 22.99 + 14.01 + (3 * 16.00) = 85.00 g/mol

Now, we can calculate the amount of NaNO3 produced using the given masses of NaOH and HNO3.

Calculation:

1. Calculate the moles of NaOH: Moles of NaOH = Mass of NaOH / Molar mass of NaOH Moles of NaOH = 20 g / 40.00 g/mol = 0.5 moles

2. Calculate the moles of HNO3: Moles of HNO3 = Mass of HNO3 / Molar mass of HNO3 Moles of HNO3 = 126 g / 63.02 g/mol = 2 moles

3. Determine the limiting reactant: The stoichiometry of the reaction shows that 2 moles of NaOH react with 1 mole of HNO3. Since there are 0.5 moles of NaOH and 2 moles of HNO3, NaOH is the limiting reactant.

4. Calculate the moles of NaNO3 produced: According to the stoichiometry, 2 moles of NaOH produce 1 mole of NaNO3. Moles of NaNO3 = 0.5 moles of NaOH * (1 mole of NaNO3 / 2 moles of NaOH) = 0.25 moles

5. Calculate the mass of NaNO3 produced: Mass of NaNO3 = Moles of NaNO3 * Molar mass of NaNO3 Mass of NaNO3 = 0.25 moles * 85.00 g/mol = 21.25 grams

The mass of NaNO3 produced in the reaction is 21.25 grams.

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