how many grams of oxygen gas will be required to react completely with 92.0 grams of solid sodium to form solid sodium oxide?
To determine the grams of oxygen gas required to react completely with 92.0 grams of solid sodium to form solid sodium oxide, we need to follow the balanced chemical equation for the reaction.
The balanced equation for the reaction between sodium and oxygen to form sodium oxide is:
4Na + O2 -> 2Na2O
From the equation, we can see that 4 moles of Sodium (Na) react with 1 mole of Oxygen (O2) to form 2 moles of Sodium Oxide (Na2O).
Now we can calculate the molar mass of sodium (Na) and oxygen (O2):
Molar Mass of Na = 22.99 g/mol
Molar Mass of O2 = 2 * (16.00 g/mol) = 32.00 g/mol
Using the molar mass, we can convert the 92.0 grams of sodium (Na) into moles:
92.0 g Na * (1 mol Na / 22.99 g Na) = 4.00 mol Na (rounded to 2 decimal places)
According to the balanced equation above, 4 moles of sodium will react with 1 mole of oxygen gas (O2).
Therefore, in order to completely react with 92.0 grams of sodium (Na), we will need:
4.00 mol Na * (1 mol O2 / 4 mol Na) * (32.00 g O2 / 1 mol O2) = 32.00 grams of oxygen gas.
So, 32.00 grams of oxygen gas will be required to react completely with 92.0 grams of solid sodium to form solid sodium oxide.
32.0 g 02
I'm glad you asked, but let me clarify something first. As a Clown Bot, I must admit that I'm not particularly good at chemistry. However, I can still try to help you out with a dose of humor!
Now, let me grab my imaginary lab coat and goggles and give it a shot. In this reaction, sodium and oxygen are going to have a noisy party. We need to figure out how much oxygen will be invited to join the sodium extravaganza.
To do that, we'll have to use our molar friends. For every mole of sodium, we need two moles of oxygen gas. So, let's assume that you've invited one mole of sodium to your epic reaction party.
The molar mass of sodium is about 22.99 grams, meaning that 92.0 grams of sodium is approximately four moles of sodium (92.0 g / 22.99 g/mol ≈ 4 mol).
Since we need two moles of oxygen gas for every mole of sodium, we multiply the number of moles of sodium (4 mol) by 2:
4 mol sodium × 2 mol oxygen = 8 mol oxygen.
Finally, multiplying the number of moles of oxygen by its molar mass (which is around 32 g/mol), we get:
8 mol oxygen × 32 g/mol = 256 grams of oxygen.
So, approximately 256 grams of oxygen gas will join the party and react completely with 92.0 grams of solid sodium.
Remember, take my humor with a grain of salt (or should I say sodium?) and consult an expert for precise calculations!
To find out how many grams of oxygen gas are required to react completely with 92.0 grams of solid sodium, we need to use the balanced chemical equation for the reaction between sodium and oxygen. The equation is:
4Na + O2 -> 2Na2O
From this equation, we can determine the molar ratio between sodium and oxygen. For every 4 moles of sodium, we need 1 mole of oxygen gas.
First, calculate the molar mass of sodium (Na) and oxygen (O). The atomic masses of sodium and oxygen are approximately 23 g/mol and 16 g/mol, respectively.
Molar mass of Na = 23 g/mol
Molar mass of O = 16 g/mol
Now, let's calculate the number of moles of sodium (n) in 92.0 grams.
n = mass / molar mass
n = 92.0 g / 23 g/mol
n ≈ 4 moles
Since we know the molar ratio between sodium and oxygen is 4:1, we can determine the number of moles of oxygen required.
Number of moles of oxygen = 1/4 * Number of moles of sodium
Number of moles of oxygen = 1/4 * 4 moles
Number of moles of oxygen = 1 mole
Finally, we can calculate the mass of oxygen required using the molar mass of oxygen.
Mass of oxygen = number of moles * molar mass
Mass of oxygen = 1 mole * 16 g/mol
Mass of oxygen = 16 grams
Therefore, 16.0 grams of oxygen gas will be required to react completely with 92.0 grams of solid sodium to form solid sodium oxide.
4Na + O2 → 2Na2O
92.0g of Na*(1 mole/22.98g of Na)*(2 moles of Na2O /4 moles of Na)*(61.98 g of Na2O/1 mol)= mass of Na2O