Given the following reaction for photosynthesis, 6CO2 + 6H2O --> C6H12O6 + 6O2, how many liters of O2 can be produced from 100 g of water at standard temperature and pressure?

Why did the scarecrow win an award? Because he was outstanding in his field! Now, let's calculate the amount of oxygen produced during photosynthesis.

To do that, we need to use stoichiometry. The balanced equation tells us that for every 6 moles of water, 6 moles of oxygen are produced.

First, let's convert the mass of water to moles. The molar mass of water (H2O) is approximately 18 g/mol. So, 100 g of water is equal to 100/18 = 5.56 moles of water.

Since the ratio of water to oxygen is 6:6, it means that for every mole of water, we get one mole of oxygen. Thus, for 5.56 moles of water, we will produce 5.56 moles of oxygen.

At standard temperature and pressure (STP), one mole of any gas occupies 22.4 liters of volume. So, 5.56 moles of oxygen will occupy 5.56 * 22.4 = 124.5 liters!

Therefore, from 100 grams of water at STP, we would expect to generate approximately 124.5 liters of oxygen. Just keep in mind that this calculation assumes the reaction goes to 100% completion, which may not always be the case. But hey, it's always good to have a little extra air, right?

To find the number of liters of O2 produced from 100 g of water during photosynthesis, we need to calculate the moles of water and then use the stoichiometry of the reaction.

1. Calculate the moles of water:
The molar mass of water (H2O) is 18.015 g/mol.
Moles of water = mass / molar mass
= 100 g / 18.015 g/mol
≈ 5.5494 mol

2. Use the stoichiometry of the reaction to determine the number of moles of O2 produced:
From the balanced equation, we can see that 6 moles of water produce 6 moles of O2.
Therefore, 1 mole of water produces 1 mole of O2.
So, the number of moles of O2 produced is also 5.5494 mol.

3. Use the ideal gas law to convert moles of O2 to liters at standard temperature and pressure (STP):
At STP, 1 mole of any gas occupies 22.4 liters.
Therefore, 5.5494 mol of O2 occupies 5.5494 * 22.4 liters.
≈ 124.20774 liters

So, approximately 124.2 liters of O2 can be produced from 100 g of water at standard temperature and pressure during photosynthesis.

To solve this problem, we need to use stoichiometry, which is the relationship between the amounts of reactants and products in a chemical reaction.

Step 1: Calculate the moles of water (H2O) in 100 g.

The molar mass of water (H2O) is 18.015 g/mol.

Number of moles = mass/molar mass
Number of moles = 100 g/18.015 g/mol ≈ 5.550 moles

Step 2: Use the balanced equation to determine the molar ratio between water (H2O) and oxygen (O2).

According to the balanced equation, for every 6 moles of water, 6 moles of oxygen are produced.

Step 3: Calculate the moles of oxygen (O2) that can be produced from the given amount of water.

Moles of oxygen (O2) = Moles of water × (6 moles O2 / 6 moles H2O)
Moles of oxygen (O2) = 5.550 moles × (6 moles O2 / 6 moles H2O) = 5.550 moles

Step 4: Convert the moles of oxygen (O2) to liters using the ideal gas law at standard temperature and pressure (STP).

At STP, the volume of 1 mole of any ideal gas is 22.4 liters.

Volume of oxygen (O2) = Moles of oxygen (O2) × 22.4 liters/mole
Volume of oxygen (O2) = 5.550 moles × 22.4 liters/mole
Volume of oxygen (O2) = 124.32 liters

Therefore, approximately 124.32 liters of O2 can be produced from 100 g of water at standard temperature and pressure.

mols H2O = grams/molar mass = ?

Convert mols H2O to mols O2. 6 mol H2O produces 6 mols O2.
1 mol O2 at stp occupies 22.4 L.

Post your work if you get stuck.