What mass of water, in grams, would have to be decomposed to produce 28.7 L of molecular oxygen at STP?
i think you can use stoichiometry for this question. i shud help
2H2O->2H2+O2
At STP, 1 mol = 22.4L
so, convert L O2-> mol O2
then, convert mol O2-> mol H20 by mole ratio
finally, multiply by molar mass H2O
To find the mass of water that needs to be decomposed to produce 28.7 L of molecular oxygen at STP (Standard Temperature and Pressure), we need to follow these steps:
Step 1: Determine the amount of oxygen gas using the ideal gas law equation:
PV = nRT
Where:
- P is the pressure, which is 1 atm at STP
- V is the volume, which is 28.7 L
- n is the number of moles of oxygen gas
- R is the ideal gas constant, which is approximately 0.0821 L·atm/(mol·K)
- T is the temperature in Kelvin, which is 273 K at STP
Rearranging the equation to solve for n (number of moles):
n = PV / RT
Substituting the given values into the equation:
n = (1 atm) * (28.7 L) / ((0.0821 L·atm/(mol·K)) * (273 K))
Calculating n, we get:
n ≈ 1.079 moles of oxygen gas
Step 2: Balanced Chemical Equation
The decomposition of water produces molecular oxygen (O₂). The balanced chemical equation for the decomposition of water is:
2 H₂O → 2 H₂ + O₂
From the equation, we can observe that 2 moles of water decompose to produce 1 mole of O₂.
Step 3: Calculate the mass of water
To calculate the mass of water, we need to use the molar mass of water, which is approximately 18.015 g/mol.
Using the molar ratio from the balanced chemical equation (2:1), we can set up the following proportion:
2 moles of H₂O / 1 mole of O₂ = x grams of H₂O / 1.079 moles of O₂
Solving for x, we get:
x ≈ (2 moles of H₂O / 1 mole of O₂) * (18.015 g/mol) * (1.079 moles of O₂)
Calculating x, we get:
x ≈ 38.94 grams of water
Therefore, approximately 38.94 grams of water would need to be decomposed to produce 28.7 L of molecular oxygen at STP.