Question

Вычислите объём водорода,который может быть получен при растворении в воде 100 г бария,если выход газа составляет 90% от теоретически возможного

Answer 1

Дано :
m (H2O) = 100г
W (вых) = 90%
Найти:
V (H2)

Решение:

100г ...................................... х
Ba + 2H2O = Ba(OH)2 + H2
137г ...................................22,4л

Ar (Ba) = 137г

Составляем пропорцию:

100г ---- х
137г ----22,4л

....... 100×22,4
х = ---------------- = 16,35 л
..............137

Ответ: V (H2) = 16,35 л

Answer 2

Calculating the Volume of Hydrogen Gas Produced

To calculate the volume of hydrogen gas that can be obtained by dissolving 100g of barium in water, given that the gas yield is 90% of the theoretically possible amount, we can use the following steps:

1. Determine the Chemical Reaction: - The reaction of barium with water can be represented as: ``` Ba + 2H2O -> Ba(OH)2 + H2 ```

2. Calculate the Amount of Hydrogen Gas Produced: - The molar mass of barium (Ba) is 137.33 g/mol. - The molar mass of hydrogen (H2) is 2.016 g/mol. - The molar ratio of barium to hydrogen in the reaction is 1:1. - The theoretical yield of hydrogen gas can be calculated using the given mass of barium and the molar mass of barium and hydrogen.

3. Apply the Yield Percentage: - The actual yield of hydrogen gas is 90% of the theoretical yield. - We can use this percentage to calculate the actual volume of hydrogen gas produced.

Calculation Steps

1. Calculate the Theoretical Yield of Hydrogen Gas: - The molar mass of barium (Ba) is 137.33 g/mol. - The molar mass of hydrogen (H2) is 2.016 g/mol. - The molar ratio of barium to hydrogen in the reaction is 1:1.

   moles of barium = mass of barium / molar mass of barium
   moles of hydrogen = moles of barium
   volume of hydrogen gas = moles of hydrogen * 22.4 L/mol (at standard conditions)
   

2. Apply the Yield Percentage: - The actual volume of hydrogen gas produced will be 90% of the calculated volume.

Calculation

Given: - Mass of barium = 100g - Yield percentage = 90%

1. Calculate the Theoretical Yield of Hydrogen Gas: - Molar mass of barium (Ba) = 137.33 g/mol - Molar mass of hydrogen (H2) = 2.016 g/mol ``` moles of barium = 100g / 137.33 g/mol = 0.729 moles moles of hydrogen = 0.729 moles volume of hydrogen gas = 0.729 moles * 22.4 L/mol = 16.34 L ```

2. Apply the Yield Percentage: - Actual volume of hydrogen gas produced = 90% * 16.34 L = 14.71 L

Conclusion

The volume of hydrogen gas that can be obtained by dissolving 100g of barium in water, with a gas yield of 90% of the theoretically possible amount, is approximately 14.71 liters.