Question

В результате реакции водорода с азотом образовалось 11.2 л аммиака (NH3). Расчитайте объёмы вступивших в реакцию веществ.​

Answer 1

Ответ:

16,8; 5,6

Объяснение:

3H₂ + N₂ = 2NH₃

V(NH₃)=11,2 л

по схеме реакции:

V(H₂)/3=V(N₂)=V(NH₃)/2

V(H₂)=3V(NH₃)/2

V(N₂)=V(NH₃)/2

V(H₂)=3*11,2/2=16,8 л

V(N₂)=11,2/2=5,6 л

Answer 2

Reaction Equation and Given Information

The reaction given is the reaction of hydrogen with nitrogen to form ammonia (NH3) and the volume of ammonia formed is 11.2 L.

Calculation of Reactant Volumes

To calculate the volumes of the substances involved in the reaction, we can use the stoichiometry of the reaction. The balanced chemical equation for the reaction of hydrogen with nitrogen to form ammonia is:

N2 + 3H2 → 2NH3

From the balanced equation, we can see that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. Therefore, the molar ratio of nitrogen to hydrogen to ammonia is 1:3:2.

Given that the volume of ammonia formed is 11.2 L, we can use the molar volume of gases at standard temperature and pressure (STP), which is 22.4 L/mol, to calculate the volumes of the substances involved.

Using the molar ratio and the volume of ammonia formed, we can calculate the volumes of the substances involved in the reaction.

Calculation

1. Calculate the moles of ammonia formed: - Moles of ammonia (NH3) = Volume of ammonia formed / Molar volume at STP - Moles of ammonia (NH3) = 11.2 L / 22.4 L/mol - Moles of ammonia (NH3) = 0.5 moles

2. Use the molar ratio to calculate the moles of hydrogen: - Moles of hydrogen (H2) = Moles of ammonia (NH3) * (3 moles of H2 / 2 moles of NH3) - Moles of hydrogen (H2) = 0.5 moles * (3 / 2) - Moles of hydrogen (H2) = 0.75 moles

3. Calculate the volume of hydrogen: - Volume of hydrogen (H2) = Moles of hydrogen (H2) * Molar volume at STP - Volume of hydrogen (H2) = 0.75 moles * 22.4 L/mol - Volume of hydrogen (H2) = 16.8 L

4. Calculate the volume of nitrogen: - Volume of nitrogen (N2) = Moles of hydrogen (H2) * (1 mole of N2 / 3 moles of H2) - Volume of nitrogen (N2) = 0.75 moles * (1 / 3) - Volume of nitrogen (N2) = 0.25 moles - Volume of nitrogen (N2) = 0.25 moles * 22.4 L/mol - Volume of nitrogen (N2) = 5.6 L

Results

- The volume of hydrogen involved in the reaction is 16.8 L. - The volume of nitrogen involved in the reaction is 5.6 L.

These calculations are based on the stoichiometry of the reaction and the given volume of ammonia formed.