If 1.50 g of H2C2O4 2H2O were heated to drive off the water of hydration, how much anhydrous H2C2O4 would remain?
Thank you! Please explain clearly.
0.34
Well, hey there! You know, when it comes to dehydrating substances, it's like giving them a fancy spa treatment - taking away all that excess water weight. So, in this case, we have 1.50g of H2C2O4 2H2O, which means it's got some water of hydration tagged along for the ride. Now, when we heat it up, we're going to say "Bye-bye" to that water, leaving us with just the anhydrous H2C2O4.
Now, to figure out how much anhydrous H2C2O4 is left, we need to consider the molar mass of H2C2O4 2H2O. You see, one mole of H2C2O4 2H2O weighs about 126.07 grams. And in that whole shebang, there's 1 mole of anhydrous H2C2O4.
So, if we take the 1.50 grams of H2C2O4 2H2O and divide it by the molar mass, we get the number of moles of H2C2O4 2H2O. Then, we multiply that number by the molar mass of anhydrous H2C2O4 to find out how much remains.
But hey, let me do the math for you, alright? *clownishly punches calculator* So, the molar mass of anhydrous H2C2O4 is about 90.04 grams per mole. Dividing 1.50g by 126.07g/mol, we get about 0.0119 moles. And multiplying that by 90.04g/mol, we find that about 1.08 grams of anhydrous H2C2O4 remain!
Voilà! After all that heating, you're left with approximately 1.08 grams of anhydrous H2C2O4.
To find the amount of anhydrous H2C2O4 that would remain after heating, we need to first understand the concept of water of hydration and its effect on the compound.
H2C2O4 2H2O is a compound known as oxalic acid dihydrate. In this compound, there are two water molecules (H2O) associated with each molecule of oxalic acid (H2C2O4). These water molecules are called water of hydration and are physically trapped within the crystal structure of the compound.
When heated, the water of hydration is driven off, leaving behind the anhydrous form of the compound, which means it no longer contains water molecules. The oxalic acid that remains after heating is called anhydrous H2C2O4.
To calculate the amount of anhydrous H2C2O4 that would remain after heating, we need to consider the molar mass of the compound and the stoichiometry between water of hydration and anhydrous oxalic acid.
The molar mass of H2C2O4 2H2O can be calculated by adding the molar masses of hydrogen (H), carbon (C), oxygen (O), and two water molecules. The molar mass values are:
H: 1.01 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
To calculate the molar mass of H2C2O4 2H2O:
(2 * H) + (2 * C) + (4 * O) + (2 * (2 * H + O))
Substituting the molar mass values:
(2 * 1.01 g/mol) + (2 * 12.01 g/mol) + (4 * 16.00 g/mol) + (2 * (2 * 1.01 g/mol + 16.00 g/mol))
This will give us the molar mass of H2C2O4 2H2O.
Next, we need to calculate the moles of H2C2O4 2H2O present in the given mass of 1.50 g using the molar mass.
Moles of H2C2O4 2H2O = Mass of compound / Molar mass of H2C2O4 2H2O
Finally, we can use the stoichiometry of the compound to determine the moles of anhydrous H2C2O4 that would be left after the water of hydration is driven off. Since there is a 1:1 stoichiometric ratio between water of hydration and anhydrous oxalic acid, the moles of H2C2O4 would be the same as the moles of H2C2O4 2H2O.
Once we have the moles of H2C2O4, we can calculate the mass of anhydrous H2C2O4 using the molar mass of H2C2O4.
Mass of anhydrous H2C2O4 = Moles of H2C2O4 * Molar mass of H2C2O4
By following these calculations, we can determine the amount of anhydrous H2C2O4 that would remain after heating 1.50 g of H2C2O4 2H2O.
H2C2O4.2H2O ==> H2C2O4 + 2H2O
mols H2C2O4.2H2O = grams/molar mass
mols H2C2O4 = same as the hydrated material.
g H2C2O4 = mols x molar mass H2C2O4.