Question

Срочнооо! Скільки грамів нітроген (V) оксиду реагує з водою, якщо утворюється 12,6 г нітратної кислоти?

Answer 1

Дано:

m(HNO3) = 12,6 г

Найти:

m(N2O5) - ?

Решение:

N2O5 + H2O => 2HNO3

n=m/M

M(HNO3) = 1 × 1 + 1 × 14 + 3 × 16 = 63 г/моль

n(HNO3) = 12,6 / 63 = 0,2 моль

Составим пропорцию, чтобы найти n(N2O5). x- количество вещества (N2O5):

x/1=0,2/2 ---> 2x=0,2 ---> x=0,1

n(N2O5) = 0,1 моль

m=n×M

M(N2O5) = 2 × 14 + 5 × 16 = 108 г/моль

m(N2O5) = 108 × 0,1 = 10,8 г

Ответ: 10,8 г

Answer 2

Calculation of Nitrogen (V) Oxide Reacting with Water

To calculate the amount of nitrogen (V) oxide that reacts with water to form 12.6 g of nitric acid, we can use the balanced chemical equation for the reaction between nitrogen (V) oxide and water.

The balanced chemical equation for the reaction is: 3NO2 + H2O → 2HNO3 + NO

From the balanced equation, we can see that 3 moles of nitrogen (V) oxide react with 1 mole of water to produce 2 moles of nitric acid and 1 mole of nitrogen monoxide.

Using the molar mass of nitrogen (V) oxide (NO2) and nitric acid (HNO3), we can calculate the amount of nitrogen (V) oxide that reacts with water.

The molar mass of nitrogen (V) oxide (NO2) is approximately 46 g/mol, and the molar mass of nitric acid (HNO3) is approximately 63 g/mol.

Calculation Steps

1. Calculate the number of moles of nitric acid formed: - Given mass of nitric acid = 12.6 g - Moles of nitric acid = Given mass / molar mass of nitric acid - Moles of nitric acid = 12.6 g / 63 g/mol - Moles of nitric acid ≈ 0.2 mol

2. Use the mole ratio from the balanced chemical equation to find the moles of nitrogen (V) oxide: - From the balanced equation, 3 moles of nitrogen (V) oxide produce 2 moles of nitric acid. - Moles of nitrogen (V) oxide = (Moles of nitric acid × 3) / 2 - Moles of nitrogen (V) oxide = (0.2 mol × 3) / 2 - Moles of nitrogen (V) oxide = 0.3 mol

3. Calculate the mass of nitrogen (V) oxide: - Mass of nitrogen (V) oxide = Moles of nitrogen (V) oxide × molar mass of nitrogen (V) oxide - Mass of nitrogen (V) oxide = 0.3 mol × 46 g/mol - Mass of nitrogen (V) oxide = 13.8 g

Answer

The amount of nitrogen (V) oxide that reacts with water to form 12.6 g of nitric acid is approximately 13.8 grams.