College chemistry help?

Question

i am a freshman at ou and I am trying to get ahead in my chemistry class so can you explain how to calculate the mass in grams of iodine (I2) that will react completely with 22.8 g of aluminum (Al) to form aluminum iodide (AlI3). thank you for any help or advice

Answer 1

number of mole for Al=22.8/27=0.84 mole. (atomic number AL=27)

1 mole AlI3 need 3 mole I

total mole I needed=3 x 0.84=2.52 mole

Atomic number Iodine=126.9

weight of Iodine needed=2.52 x 126.9=319.8g.

Answer 2

3 atoms of iodine react with 1 atom of aluminum.

The atomic masses of iodine and aluminum are 27 and 127, so

381g of iodine react with 27 g of aluminum

381 x 22.8/27 = 321g of iodine react with 22.8 g aluminum

Answer 3

Always start by writing the complete balanced chemical equation:

2Al + 3I2 --> 2AlI3

I'm going to show you two methods. Both methods are easy. The first method takes advantage of the idea that in a chemical reaction mass is conserved, so the mass of the product should be the total mass of all of the reagents. We know the mass of one of the reagents (Al) and assuming this reagent is the limiting reagent we can find the theoretical mass of the product. The mass of the other reagent (I2) can be determined by subtracting the mass of Al from the mass of AlI3.

The second method just involves using the stoichiometry of the reaction and dimensional analysis.

Method 1:

To find the mass of AlI3 that is produced find the moles of Al and relate the moles of AlI3 to the moles of Al. Since there are 2 moles of AlI3 per 2 moles of Al the ratio is 1:1. Use the molar mass of AlI3 to find the mass of AlI3. The molar mass of Al is 26.982 g/mol and the molar mass of AlI3 is 407.682 g/mol.

(22.8 g Al)(1 mol Al/ 26.982 g Al)(2 mol AlI3/ 2 mol Al)(407.682 g AlI3/ 1 mol AlI3) = 344.49 g AlI3

The mass of I2 is:

Al + I2 = AlI3

I2 = AlI3 - Al = 344.49 - 22.8 = 321.69 g I2

Method 2:

Using stoichiometry, find the moles of AlI3 and relate the moles of I2 to the moles of AlI3 and use the molar mass of I2 (253.8 g/mol) to find the mass of I2. There are 3 moles of I2 per 2 moles of AlI3 so the ratio is 3:2.

(344.49 g AlI3)(1 mol AlI3/ 407.682 g AlI3)(3 mol I2/ 2 mol AlI3)( 253.8 g I2/ mol I2) = 321.69

The mass of I2 is the same for both methods which means that I did something right.

Notice that the mass of I2 is much larger than the mass of Al, so I was correct to assume that Al is the limiting reagent.

You can't go wrong with either of these methods. I hope I explained them both well enough for you. Good luck!