Assuming all volume measurements are made

at the same temperature and pressure, what
volume of hydrogen gas is needed to react
completely with 3.25 L of oxygen gas to produce
water vapor?
Answer in units of L.

See your problems above. Same procedure for all gas problems.

To determine the volume of hydrogen gas needed to react completely with 3.25 L of oxygen gas, we need to use the balanced chemical equation for the reaction.

The balanced chemical equation for the reaction between hydrogen gas (H2) and oxygen gas (O2) to produce water vapor (H2O) is:

2H2 + O2 -> 2H2O

The stoichiometry of the balanced equation tells us that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water vapor.

1 mole of any gas occupies 22.4 L at standard temperature and pressure (STP).

To calculate the volume of hydrogen gas, we need to find the ratio of moles of hydrogen to moles of oxygen, and then convert the moles of hydrogen gas into liters.

Step 1: Convert the given volume of oxygen gas to moles.
Using the ideal gas law, at STP, 1 mole of any gas occupies 22.4 L. Therefore, we can use the following conversion:

3.25 L O2 * (1 mole O2 / 22.4 L O2) = 0.145 moles O2

So, we have 0.145 moles of oxygen gas.

Step 2: Determine the moles of hydrogen gas.
Using the stoichiometry of the balanced equation, we can determine the mole ratio between hydrogen and oxygen. From the balanced equation, we know that 2 moles of hydrogen gas react with 1 mole of oxygen gas.

Therefore, if we have 0.145 moles of oxygen gas, we need 2 * 0.145 = 0.29 moles of hydrogen gas to react completely.

Step 3: Convert moles of hydrogen gas to volume.
Using the ideal gas law at STP, we can convert the moles of hydrogen gas to volume in liters:

0.29 moles H2 * (22.4 L / 1 mole H2) = 6.496 L

Therefore, approximately 6.496 L of hydrogen gas is needed to react completely with 3.25 L of oxygen gas to produce water vapor.