Hydrogen burns to give water. If 200cm3 of H2 reacts with 150cm3 of O2 what volume of water vapor is produced
2H2 + O2 ==> 2H2O
When working with gases one need not go through mols but may use volume directly.
Convert 200 L H2 to H2O using the coefficients in the balanced equation.
200 cc H2 x (2 mols H2O/2 mol H2) = 200 cc H2O
Convert 150 cc O2 to H2O.
150 cc O2 x (2 mols H2O/1 mol O2) = 3-- cc H2O.
You note that these numbers don't agree which means one of them is wrong. The correct value in limiting reagent problems (and this is a limiting reagent problem because amounts are given for BOTH reactants) is ALWAYS the smaller value. Therefore, you can produce 200 cc H2O this way.
To determine the volume of water vapor produced when hydrogen reacts with oxygen, we need to use the balanced chemical equation for the reaction and apply the principles of stoichiometry.
The balanced equation for the reaction between hydrogen (H2) and oxygen (O2) to form water (H2O) is:
2H2 + O2 -> 2H2O
According to the stoichiometry of the reaction, it indicates that 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O. Since the volume of a gas is directly proportional to the number of moles, we can solve for the volume of water vapor produced.
First, we need to convert the given volumes of hydrogen and oxygen into moles using the ideal gas law:
V = nRT/P
Where:
V = volume of gas (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin
P = pressure in atm (assuming constant)
Assuming constant temperature and pressure, we can solve for the number of moles of H2 and O2:
n(H2) = (200 cm^3) / (1000 cm^3/L) = 0.2 L
n(O2) = (150 cm^3) / (1000 cm^3/L) = 0.15 L
Now, we compare the moles of H2 to O2 ratios in the given reaction:
H2:O2 = 2:1
Since the ratio indicates that 2 moles of H2 react with 1 mole of O2, and we have 0.2 moles of H2 and 0.15 moles of O2, we can determine the limiting reactant:
O2 is limiting, as it is less than half the amount needed to fully react with the given amount of H2.
Now, we can use the stoichiometry to determine the moles of water vapor produced:
n(H2O) = 2 * n(O2) = 2 * 0.15 = 0.3 moles
Finally, we can convert the moles of water vapor back into volume:
V(H2O) = n(H2O) * V_molar
Where V_molar is the molar volume of gas at the given conditions.
Assuming constant temperature and pressure (25 degrees Celsius and 1 atm), the molar volume is approximately 22.4 L/mol.
V(H2O) = 0.3 moles * 22.4 L/mol = 6.72 L
Therefore, the volume of water vapor produced in this reaction is approximately 6.72 liters.
To determine the volume of water vapor produced when hydrogen reacts with oxygen, we need to use the balanced equation for the combustion of hydrogen:
2H2 + O2 -> 2H2O
The balanced equation tells us that 2 moles of hydrogen gas (H2) react with 1 mole of oxygen gas (O2) to produce 2 moles of water vapor (H2O). From here, we can use the concept of mole ratios to calculate the volume of water vapor produced.
Step 1: Convert the given volumes of hydrogen and oxygen to moles.
To convert from volume to moles, we need to use the ideal gas law equation, which states: PV = nRT. However, because the volume is given at the same temperature and pressure, we can simplify the equation to V = n.
Given:
Volume of H2 = 200 cm3
Volume of O2 = 150 cm3
Assuming that the hydrogen and oxygen are measured under the same conditions:
n(H2) = V(H2) / 22.4 (since 1 mole of any gas occupies 22.4 liters, or 22,400 cm3, at standard temperature and pressure conditions)
n(H2) = 200 cm3 / 22.4 cm3/mol = 8.93 moles (rounded to two decimal places)
n(O2) = V(O2) / 22.4
n(O2) = 150 cm3 / 22.4 cm3/mol = 6.70 moles (rounded to two decimal places)
Step 2: Determine the limiting reactant.
To determine the limiting reactant, we compare the mole ratios of hydrogen to oxygen in the balanced equation. The reactant that is not present in the required stoichiometric quantity is the limiting reactant.
According to the balanced equation, 2 moles of H2 react with 1 mole of O2. So, if we have 8.93 moles of H2, we would need 8.93 / 2 = 4.46 moles of O2 for complete reaction.
Since we have only 6.70 moles of O2, which is larger than the required 4.46 moles, O2 is in excess, and H2 is the limiting reactant.
Step 3: Calculate the moles of water vapor produced.
As stated in the balanced equation, 2 moles of H2 react to produce 2 moles of H2O. Since H2 is the limiting reactant, we can conclude that the moles of water vapor produced will be equal to the moles of H2, which is 8.93 moles.
Step 4: Convert the moles of water vapor to volume.
Using the ideal gas law equation (V = n), we can convert moles of water vapor to volume.
n(H2O) = n(H2) = 8.93 moles
V(H2O) = n(H2O) * 22.4 cm3/mol
V(H2O) = 8.93 moles * 22.4 cm3/mol = 199.83 cm3 (rounded to two decimal places)
Therefore, the volume of water vapor produced is approximately 199.83 cm3.