Assuming all volume measurements are made
at the same temperature and pressure, what
volume of hydrogen gas is needed to react
completely with 3.75 L of oxygen gas to produce
water vapor?
2H2 + O2 --> 2H2O
When using gases, one may take a shortcut and use volume as if it were mols.
3.75L O2 x (2 mols H2/1 mol O2) = 3.75 x 2/1 = ? L H2.
To answer this question, we need to use the balanced chemical equation for the reaction between hydrogen gas (H2) and oxygen gas (O2) to produce water vapor (H2O). The balanced equation is:
2H2(g) + O2(g) -> 2H2O(g)
From the balanced equation, we can see that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water vapor.
1. First, determine the number of moles of oxygen gas (O2) using the ideal gas law equation: PV = nRT.
Given:
- Volume of oxygen gas (V) = 3.75 L
- Assuming temperature (T) and pressure (P) are constant
2. Rearrange the ideal gas law equation to solve for the number of moles (n) of oxygen gas:
n = PV / RT
3. Plug in the values:
n = (Pressure of oxygen gas * Volume of oxygen gas) / (Ideal gas constant * Temperature)
Note: The ideal gas constant (R) is a constant value of 0.0821 L·atm/(mol·K).
4. Calculate the value of n.
Next, we can use the stoichiometry of the balanced equation to find the volume of hydrogen gas (H2) needed:
5. Determine the mole ratio of hydrogen gas (H2) to oxygen gas (O2) from the balanced equation. In this case, the ratio is 2:1.
6. Calculate the number of moles of hydrogen gas using the mole ratio from step 5.
7. Finally, use the ideal gas law equation to find the volume of hydrogen gas (V) corresponding to the calculated moles of hydrogen gas.
That will give you the volume of hydrogen gas needed to react completely with the given amount of oxygen gas to produce water vapor.