Question

Какой объем воздуха потребуется для сжигания 28л водорода? Объемная доля кислорода в воздухе составляет 21%. Объемы газов измерены при нормальных условиях.

Answer 1

2H2 + O2 = 2H2O

По уравнению реакции видим, что в реакции участвуют 2 моль водорода и 1 моль кислорода. 1 моль любого газа занимает объем 22.4 л. Значит, если водорода взяли 28 л, то кислорода потребуется в два раза меньше - 14 л.

 
Объемная доля кислорода в воздухе составляет 21%.  Следовательно, V(воздуха) = 14*100/21 = 66.7 л

Answer 2

Calculating the Volume of Air Needed to Burn 28L of Hydrogen

To calculate the volume of air needed to burn 28L of hydrogen, we can use the fact that the stoichiometric ratio of hydrogen to oxygen in the combustion reaction is 2:1. This means that for every 2 volumes of hydrogen, 1 volume of oxygen is needed. Since air is approximately 21% oxygen, we can calculate the volume of air needed.

Step 1: Calculate the Volume of Oxygen Needed - Given: 28L of hydrogen - Stoichiometric ratio of hydrogen to oxygen: 2:1

Using the stoichiometric ratio, we can calculate the volume of oxygen needed: 28L of hydrogen * (1 volume of oxygen / 2 volumes of hydrogen) = 14L of oxygen.

Step 2: Calculate the Volume of Air Needed - Since air is approximately 21% oxygen, we can calculate the volume of air needed using the volume of oxygen calculated in Step 1.

To find the volume of air needed, we can use the following formula: Volume of air = Volume of oxygen / 0.21

Substituting the value of the volume of oxygen: Volume of air = 14L / 0.21 = 66.67L

Therefore, the volume of air needed to burn 28L of hydrogen is approximately 66.67L.

Conclusion

To burn 28L of hydrogen, approximately 66.67L of air, which contains the required amount of oxygen, is needed.