Question

Сколько килограммов 68%-й азотной кислоты можно получить из 276 кг (н.у.) оксида азота (IV)?

Answer 1

4NО2+О2+2Н2О=4НNО3 
n(NО2)=m/M=27600/46=6000моль 
n(НNО3)=6000 
m(HNO3)=6000*63=378 кг 
m(HNO3)раствора=378/0,68=556 кг

Answer 2

Calculating the Amount of Nitric Acid

To calculate the amount of nitric acid that can be obtained from 276 kg of nitrogen(IV) oxide, we can use the following chemical equation:

3N2O + H2O → 4HNO3

According to the equation, 3 moles of nitrogen(IV) oxide react with water to produce 4 moles of nitric acid.

Calculation Steps

1. Convert the mass of N2O to moles: - The molar mass of N2O is approximately 44.013 g/mol. - Calculate the number of moles in 276 kg (276,000 g) of N2O.

2. Determine the moles of HNO3 produced: - Use the mole ratio from the balanced chemical equation to find the moles of HNO3 produced.

3. Convert moles of HNO3 to kilograms: - Calculate the mass of HNO3 produced from the moles obtained.

Calculation

1. Convert the mass of N2O to moles: - 276,000 g / 44.013 g/mol ≈ 6270 moles of N2O

2. Determine the moles of HNO3 produced: - Using the mole ratio from the balanced chemical equation: - 3 moles of N2O produce 4 moles of HNO3 - (6270 moles N2O * 4 moles HNO3) / 3 moles N2O ≈ 8360 moles of HNO3

3. Convert moles of HNO3 to kilograms: - The molar mass of HNO3 is approximately 63.012 g/mol. - 8360 moles * 63.012 g/mol ≈ 526,000 g ≈ 526 kg of HNO3

Answer

Therefore, approximately 526 kilograms of 68%-concentrated nitric acid can be obtained from 276 kilograms of nitrogen(IV) oxide.