How many grams of Al2O3 are produced when 50.3g of Al reacts?
Well, it's time to do a little bit of chemistry math, but fear not, I'll be your comedic guide through this numerical journey! Let's break it down.
The reaction equation tells us that:
2Al + 3O2 -> 2Al2O3
Now, the molar mass of Al2O3 is approximately 101.96 g/mol. To find out how many moles of Al2O3 are produced from 50.3g of Al, we need to do a little bit of molar magic.
First, we need to figure out the molar mass of Al, which is 26.98 g/mol. Then we divide the mass of Al given (50.3g) by its molar mass:
50.3g / 26.98 g/mol = approximately 1.866 mol
Now, since the ratio of Al to Al2O3 is 2:2, this means that for every 2 moles of Al, we get 2 moles of Al2O3. So if we look at the ratio, we can conclude that 1.866 moles of Al will give us 1.866 moles of Al2O3.
Finally, to find the mass of Al2O3 produced, we multiply the number of moles of Al2O3 by its molar mass:
1.866 mol * 101.96 g/mol = approximately 190.07g
Therefore, when 50.3g of Al reacts, approximately 190.07g of Al2O3 are produced. Voilà!
I hope my comedic twist on this chemistry calculation added a bit of laughter to your day!
To determine the number of grams of Al₂O₃ produced when 50.3g of Al reacts, we need to know the balanced chemical equation for the reaction. Without the equation, it is not possible to provide a step-by-step solution.
Please provide the balanced equation and any additional information if available.
To determine the number of grams of Al2O3 produced when 50.3g of Al reacts, we need to use the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction involving Al and Al2O3 is:
2Al + 3O2 -> 2Al2O3
From the equation, we can see that 2 moles of Al react to produce 2 moles of Al2O3. Now, we need to convert the grams of Al to moles using the molar mass of Al.
The molar mass of Al is 26.98g/mol.
To find the number of moles of Al in 50.3g of Al, we use the formula:
moles = mass / molar mass
moles = 50.3g / 26.98g/mol ≈ 1.8653 moles
Now, from the balanced equation, we know that 2 moles of Al react to produce 2 moles of Al2O3. Therefore, 1 mole of Al will produce 1 mole of Al2O3.
So, the number of moles of Al2O3 produced is also approximately 1.8653 moles.
Finally, we need to convert the moles of Al2O3 to grams using the molar mass of Al2O3.
The molar mass of Al2O3 is 101.96g/mol.
grams = moles * molar mass
grams = 1.8653 moles * 101.96g/mol ≈ 190.11g
Therefore, approximately 190.11 grams of Al2O3 are produced when 50.3 grams of Al reacts.
4 Al + 2 O3 -> 2 Al2O3
find the moles of Al ... (sample mass) / (molar mass)
find moles of O reacted ... three moles of O for every two moles of Al
... (3/2) * (moles Al) = (moles O)
find mass of O reacted ... (moles O) * (molar mass O)
add the mass of O to the mass of Al