2H2O --> 2H2 + O2
if 30 mL of hydrogen in the above reaction, how many milliliters of oxygen are produced?
Well, isn't it quite generous of hydrogen to give oxygen so much space to stretch its lungs? In the reaction 2H2O → 2H2 + O2, it seems that every 2 molecules of water (H2O) produce 2 molecules of hydrogen (H2) and just 1 molecule of oxygen (O2). So, if we have 30 mL of hydrogen, that means we need half the amount of oxygen. Hence, we would get a rather concise 15 mL of oxygen. It's all about sharing!
According to the balanced equation: 2H2O --> 2H2 + O2
The stoichiometry of the reaction states that for every 2 molecules of H2O, 1 molecule of O2 is produced.
Since the equation is balanced in terms of moles, we will need to convert the given volume of hydrogen gas into moles.
To do this, we need to know the density of hydrogen gas at the given conditions. The density of hydrogen gas at standard temperature and pressure (STP) is about 0.0899 grams per liter (g/L), or 0.0899 g/mL.
Given that the volume of hydrogen gas is 30 mL, we can calculate the mass of hydrogen gas using the density:
Mass of hydrogen = volume of hydrogen gas × density of hydrogen
Mass of hydrogen = 30 mL × 0.0899 g/mL
Mass of hydrogen = 2.697 g
The molar mass of hydrogen (H2) is approximately 2 g/mol. Therefore, the number of moles of hydrogen can be calculated as follows:
Number of moles of hydrogen = mass of hydrogen / molar mass of hydrogen
Number of moles of hydrogen = 2.697 g / 2 g/mol
Number of moles of hydrogen = 1.3485 mol
From the balanced equation, we know that 2 moles of hydrogen gas (H2) produces 1 mole of oxygen gas (O2).
Therefore, the number of moles of oxygen produced will be half the number of moles of hydrogen consumed.
Number of moles of oxygen = 1.3485 mol / 2
Number of moles of oxygen = 0.67425 mol
Now, we need to convert the number of moles of oxygen gas into milliliters.
Again, we need to know the density of oxygen gas at the given conditions. The density of oxygen gas at standard temperature and pressure (STP) is about 1.429 grams per liter (g/L), or 1.429 g/mL.
We can calculate the volume of oxygen gas using the following equation:
Volume of oxygen gas = number of moles of oxygen × molar volume
Volume of oxygen gas = 0.67425 mol × 22.4 L/mol (molar volume at STP)
Volume of oxygen gas = 15.11 L
Finally, we can convert the volume of oxygen gas from liters to milliliters:
Volume of oxygen gas = 15.11 L × 1000 mL/L
Volume of oxygen gas = 15110 mL
Therefore, if 30 mL of hydrogen gas is consumed in the reaction, approximately 15110 mL of oxygen gas will be produced.
To determine how many milliliters of oxygen are produced in the given reaction, we need to use the stoichiometry of the reaction. The balanced equation tells us that for every 2 moles of water (H2O), we produce 2 moles of hydrogen gas (H2) and 1 mole of oxygen gas (O2).
Since we're given the volume of hydrogen gas (30 mL), we need to convert this volume to moles using the ideal gas law equation:
PV = nRT
Where:
P = pressure (assumed constant)
V = volume
n = number of moles
R = ideal gas constant
T = temperature (assumed constant)
Since the pressure and temperature are not specified, we can assume they remain constant. The ideal gas constant (R) is also a constant value. Therefore, we can write the equation to solve for moles as:
n = (PV) / RT
Now, we need to convert the moles of hydrogen gas (H2) to moles of oxygen gas (O2) using the stoichiometry ratio of 2:1. This means that for every 2 moles of hydrogen gas, we produce 1 mole of oxygen gas.
Given the mole ratio, we can now determine the moles of oxygen gas produced.
Once we have the moles of oxygen gas produced, we can convert it back to volume using the ideal gas law equation or the relationship that 1 mole of any gas at STP (Standard Temperature and Pressure) occupies 22.4 L (or 22,400 mL).
Therefore, we have:
1. Convert the volume of hydrogen gas to moles using the ideal gas law equation.
2. Use the stoichiometry ratio to determine the moles of oxygen gas produced.
3. Convert the moles of oxygen gas back to volume using the ideal gas law equation or converting moles to milliliters at STP.
By following these steps, you will be able to calculate the volume of oxygen gas produced in milliliters.