Question

Даю 40 баллов! Определите массу воды, которая образовалась при взаимодействии 80л водорода и кислорода. Пожалуйста, с подробным решением, по формулам, 8 класс

Answer 1

Ответ:

64,3 г

Объяснение:

Дано:

V(H₂)=80л

Найти:

m(H₂O)-?

Решение:

Составим уравнение реакции

2H₂+O₂=2H₂O

Рассчитаем количество водорода:

n(H₂)=V(H₂)/Vm=80/22,4=3,57 моль

По уравнению рассчитаем количество воды:

n(H₂O)=1/1*n(H₂)=3,57 моль

Найдём массу воды:

m(H₂O)=n(H₂O)*M(H₂O)=3,57*18=64,3 г

Answer 2

Calculating the Mass of Water Formed from the Reaction of Hydrogen and Oxygen

To determine the mass of water formed from the reaction of 80 liters of hydrogen and oxygen, we can use the balanced chemical equation for the formation of water from hydrogen and oxygen.

The balanced chemical equation for the formation of water from hydrogen and oxygen is:

2H2 + O2 → 2H2O

This equation shows that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.

Step-by-Step Calculation

1. Convert Liters to Moles: - Given: 80 liters of hydrogen and oxygen - We need to convert the volume of gases to moles using the ideal gas law, which states that 1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 liters.

- For Hydrogen (H2): - Number of moles of hydrogen = Volume of hydrogen / Molar volume at STP - Number of moles of hydrogen = 80 L / 22.4 L/mol - Number of moles of hydrogen = 3.57 moles

- For Oxygen (O2): - Number of moles of oxygen = Volume of oxygen / Molar volume at STP - Number of moles of oxygen = 80 L / 22.4 L/mol - Number of moles of oxygen = 3.57 moles

2. Determine Limiting Reactant: - To find the limiting reactant, we compare the moles of each reactant to the coefficients in the balanced chemical equation.

- In this case, both hydrogen and oxygen have the same number of moles, so the limiting reactant is either one.

3. Calculate the Mass of Water Formed: - Since the balanced chemical equation shows that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water, we can use the mole ratio to calculate the moles of water formed.

- Moles of water formed = Moles of limiting reactant (hydrogen or oxygen) / Coefficient of water in the balanced equation - Moles of water formed = 3.57 moles / 2 - Moles of water formed = 1.79 moles

4. Convert Moles to Grams: - Finally, we can convert the moles of water formed to grams using the molar mass of water.

- Molar mass of water (H2O) = 2(1.008 g/mol) + 15.999 g/mol - Molar mass of water (H2O) = 18.015 g/mol

- Mass of water formed = Moles of water formed × Molar mass of water - Mass of water formed = 1.79 moles × 18.015 g/mol - Mass of water formed ≈ 32.26 grams

Conclusion:

The mass of water formed when 80 liters of hydrogen and oxygen react is approximately 32.26 grams.