Given the equation:
6CO2(g) + 6H2O(l) -> C6H12O6(s) + 6O2(g)
What is the minimum number of liters of CO2(g), measured at STP, needed to produce 32.0 grams of oxygen?
1) 264 L
2) 32.0 L
3) 192 L
4) 22.4 L
How do I solve?
22.4
Well, this question seems like a gas to me! Let's break it down.
From the balanced equation, we can see that for every 6 moles of CO2, we get 6 moles of O2. So the mole ratio is 1:1. This means that the number of moles of CO2 needed to produce a certain amount of O2 will be the same.
We can use the molar mass of O2 to convert grams to moles. The molar mass of O2 is approximately 32 g/mol. So, 32.0 grams of O2 is equal to 1 mole of O2.
Since the mole ratio of CO2 to O2 is 1:1, we need 1 mole of CO2 to produce 1 mole of O2. And we know that at STP, 1 mole of any gas occupies 22.4 liters.
So, the minimum number of liters of CO2 needed to produce 32.0 grams of oxygen at STP is 22.4 liters. Therefore, the correct answer is option 4) 22.4 L.
Now that's what I call a "lit-"tle bit of gas math!
To solve this problem, we can use the balanced chemical equation and convert the given mass of oxygen to the volume of carbon dioxide (CO2) at Standard Temperature and Pressure (STP).
Step 1: Determine the molar mass of oxygen (O2)
Oxygen has a molar mass of 32.0 grams/mol since its atomic mass is 16.0 g/mol.
Step 2: Convert the mass of oxygen (O2) to moles
Divide the given mass of oxygen by the molar mass to convert grams to moles:
moles of O2 = mass of O2 / molar mass of O2
moles of O2 = 32.0 g / 32.0 g/mol
moles of O2 = 1.0 mol
Step 3: Use the stoichiometry of the balanced equation to convert moles of O2 to moles of CO2
From the balanced equation, we can see that 6 moles of CO2 are produced for every 6 moles of O2. Therefore, the ratio is 1:1.
So the moles of CO2 required will also be 1.0 mol.
Step 4: Convert moles of CO2 to volume at STP
At STP, 1 mole of any gas occupies 22.4 liters of volume. Therefore, 1.0 mol of CO2 will occupy 22.4 liters.
Therefore, the minimum number of liters of CO2 needed to produce 32.0 grams of oxygen is 22.4 L.
Hence, the correct answer is 4) 22.4 L.
To determine the minimum number of liters of CO2(g) needed to produce 32.0 grams of oxygen, we need to use stoichiometry - the study of the quantitative relationships between reactants and products in a chemical reaction.
First, let's calculate the number of moles of oxygen that are produced using the given mass:
Mass of oxygen = 32.0 grams
Molar mass of oxygen (O2) = 32.0 g/mol
Number of moles of oxygen = Mass / Molar mass = 32.0 g / 32.0 g/mol = 1 mol
According to the balanced equation, the stoichiometric ratio between CO2 and O2 is 6:6, which means for every 6 moles of CO2, 6 moles of O2 are produced.
Now, we can use this stoichiometric ratio to calculate the number of moles of CO2 required to produce 1 mole of O2:
Number of moles of CO2 = Number of moles of O2 / Stoichiometric ratio = 1 mol / 6 = 1/6 mol
At STP (Standard Temperature and Pressure), one mole of any ideal gas occupies a volume of 22.4 liters. Therefore, the minimum number of liters of CO2(g) needed can be calculated as follows:
Number of liters of CO2(g) = Number of moles of CO2 x 22.4 L/mol = (1/6) mol x 22.4 L/mol = 3.73 L
Since the minimum number of liters of CO2(g) needed is approximately 3.73 L, the closest answer choice is 4) 22.4 L, since it is the only option that is in the correct range.
32.0 g O2 = 1 mole O2
1 mole O2 = 22.4 L
The mole ratio shown by the coefficients in the chemical equation is the same as the volume (liters) ratio. So,
6 L of CO2 produce 6 L O2, or
1 L CO2 make 1 L O2
How many liters CO2 would produce 22.4 L O2 ??