hydrogen gas reacts with oxygen gas to produce water if a student fills a 250-mL flask with hydrogen at STP, what volume of oxygen is necessary for the reaction?
To determine the volume of oxygen necessary for the reaction, we can use the balanced chemical equation for the reaction between hydrogen and oxygen to produce water:
2H₂ + O₂ → 2H₂O
According to the stoichiometry of the equation, it shows that two moles of hydrogen will react with one mole of oxygen to produce two moles of water. Therefore, the ratio between the volume of hydrogen and oxygen should also be 2:1.
Given that the student fills a 250-mL flask with hydrogen at STP, we need to calculate the volume of oxygen required.
First, we convert the volume of hydrogen (in mL) to moles using the ideal gas law equation:
PV = nRT
Since the conditions are STP (standard temperature and pressure), we can use the following values:
- Temperature (T) = 273.15 K
- Pressure (P) = 1 atm
- Volume (V) = 250 mL (which we'll convert to liters)
Converting the volume to liters:
250 mL = 250/1000 = 0.25 L
Now, rearrange the ideal gas law equation to solve for moles (n):
n = PV / RT
Substituting the known values:
n = (1 atm * 0.25 L) / (0.0821 L·atm/(mol·K) * 273.15 K)
n ≈ 0.112 mol
According to the stoichiometry of the reaction, 2 moles of hydrogen react with 1 mole of oxygen. Therefore, since we have 0.112 mol of hydrogen, we need half that amount of oxygen:
0.112 mol hydrogen * (1 mol oxygen / 2 mol hydrogen) ≈ 0.056 mol oxygen
Finally, we can convert the moles of oxygen back to volume by using the ideal gas law equation:
V = nRT / P
Substituting the known values:
V = (0.056 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm
V ≈ 1.16 L
Therefore, the volume of oxygen necessary for the reaction is approximately 1.16 liters.
You need to find the caps key on your computer and use it. While you are at it, find the period also.
2H2 + O2 ==> 2H2O
When working with gases one can use volume interchangeably as moles. Therefore, If you have 250 mL H2 you will need 250 x 1 mol (O2/2 mols H2) = 125 mL O2 at the same T and P.