Given the following equation : 2Al+6HCl 2AlCl 3 +3H 2 If 3.00 grams of H 2 were produced, how many grams of Al reacted ? (Molar mass of Al = 26.98g / (mol) , molar mass of H 2 = 2.016 g/mol)
26.8
Well, it seems like Al decided to put on some extra weight before the reaction. But don't worry, I'm here to help you solve this equation!
Let's start by using the given molar mass of H2 to convert the grams into moles. We have 3 grams of H2, so that would be 3 g / 2.016 g/mol = 1.49 moles of H2.
Now, looking at the balanced equation, we can see that there is a 2:3 ratio between Al and H2. This means that for every 2 moles of Al, we get 3 moles of H2.
Since we found that we have 1.49 moles of H2, we can use the ratio to find the moles of Al. Setting up a proportion:
2 moles Al / 3 moles H2 = x moles Al / 1.49 moles H2
Cross multiplying, we get:
3 moles H2 * x moles Al = 2 moles Al * 1.49 moles H2
Simplifying this, we have:
3x = 2 * 1.49
3x = 2.98
Dividing both sides by 3, we find:
x = 0.99
So, approximately 0.99 moles of Al reacted. Now, we can convert moles of Al into grams:
0.99 moles Al * 26.98 g/mol = 26.67 grams of Al (approximately).
Therefore, about 26.67 grams of Al reacted. Keep in mind that this is an approximate value. Now, go tell Al to lay off the extra weight next time!
To find the number of grams of Al that reacted, we need to use stoichiometry and the molar ratio between Al and H2.
First, let's calculate the number of moles of H2 produced using the given mass and molar mass of H2:
Mass of H2 = 3.00 grams
Molar mass of H2 = 2.016 g/mol
Number of moles of H2 = Mass of H2 / Molar mass of H2
= 3.00 g / 2.016 g/mol
≈ 1.49 mol H2
Now, let's set up the stoichiometric ratio between Al and H2 from the balanced chemical equation:
2Al + 6HCl -> 2AlCl3 + 3H2
From the equation, we can see that 2 moles of Al react to produce 3 moles of H2. So, the molar ratio of Al to H2 is 2:3.
Now, we can use this ratio to find the number of moles of Al that reacted:
Number of moles of Al = (Number of moles of H2) * (Molar ratio of Al to H2)
= 1.49 mol * (2 mol Al / 3 mol H2)
≈ 0.993 mol Al
Finally, we can find the mass of Al that reacted using the molar mass of Al:
Mass of Al = Number of moles of Al * Molar mass of Al
= 0.993 mol * 26.98 g/mol
≈ 26.9 grams
Therefore, approximately 26.9 grams of Al reacted.
i got 3.78
13.1
I have rewritten the equation to make it easier to read and understand.
2Al + 6HCl ==> 2AlCl3 + 3H2
How many mols H2 are in 3 g H2? That's mols = grams/molar mass = 3.00/2 = 1.5.
Convert 1.5 mols H2 to mols Al. That's done by using the coefficients in the balanced equation; i.e., 1.5 mols H2 x (2 mol Al/3 mol H2) = 1.5 x 2/3 = ?
Now convert mols Al to grams. g Al = mols Al x atomic mass Al = ?
Post your work if you get stuck.