Answer:
Step-by-step explanation:
The balanced chemical equation for the reaction in which water (H2O) is broken down into hydrogen gas (H2) and oxygen gas (O2) is:
2 H2O → 2 H2 + O2
According to the stoichiometry of this equation, for every 2 moles of water that react, 1 mole of oxygen gas is consumed. Therefore, to determine how many moles of oxygen gas were consumed, we need to find how many moles of water reacted.
Since we know that 0.25 moles of water were produced, we can assume that the same number of moles of water reacted, which means that:
0.25 moles H2O reacted
Using the stoichiometry of the balanced equation, we can see that for every 2 moles of water that react, 1 mole of oxygen gas is consumed. Therefore, the number of moles of oxygen gas consumed can be calculated as:
0.25 moles H2O x (1 mole O2 / 2 moles H2O) = 0.125 moles O2
Finally, we can use the ideal gas law, PV = nRT, to calculate the volume of oxygen gas consumed, assuming standard temperature and pressure (STP) of 0°C (273 K) and 1 atmosphere (1 atm):
V = nRT/P
V = (0.125 mol)(0.0821 L·atm/mol·K)(273 K)/(1 atm)
V = 2.7 L
Therefore, 2.7 liters of oxygen gas were consumed in the reaction.The balanced chemical equation for the reaction in which water (H2O) is broken down into hydrogen gas (H2) and oxygen gas (O2) is:
2 H2O → 2 H2 + O2
According to the stoichiometry of this equation, for every 2 moles of water that react, 1 mole of oxygen gas is consumed. Therefore, to determine how many moles of oxygen gas were consumed, we need to find how many moles of water reacted.
Since we know that 0.25 moles of water were produced, we can assume that the same number of moles of water reacted, which means that:
0.25 moles H2O reacted
Using the stoichiometry of the balanced equation, we can see that for every 2 moles of water that react, 1 mole of oxygen gas is consumed. Therefore, the number of moles of oxygen gas consumed can be calculated as:
0.25 moles H2O x (1 mole O2 / 2 moles H2O) = 0.125 moles O2
Finally, we can use the ideal gas law, PV = nRT, to calculate the volume of oxygen gas consumed, assuming standard temperature and pressure (STP) of 0°C (273 K) and 1 atmosphere (1 atm):
V = nRT/P
V = (0.125 mol)(0.0821 L·atm/mol·K)(273 K)/(1 atm)
V = 2.7 L
Therefore, 2.7 liters of oxygen gas were consumed in the reaction.