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bagirrra123 [75]
3 years ago
12

Many grams of aluminum are required to produce 3.5 moles Al2O3 in the presence of excess O2?

Chemistry
1 answer:
Aleks [24]3 years ago
4 0
The  grams  of aluminum  that are required   to produce  3.5  moles of AlO3  in  presence of excess O2   is calculated as  below

write  the  equation for reaction
4 Al + 3O2 =2 Al2O3

by use of mole  ratio between  Al  to  Al2O3   which  is  4 :2  the moles of  Al 
=3.5 x4/2 = 7  moles

mass of Al  =  moles /   x molar mass

= 7 moles  x27 g/mol  =189  grams

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A. Invisible

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For which of the following aqueous solutions would one expect to have the largest van’t Hoff factor (i)? a. 0.050 m NaCl b. 0.50
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The van't hoff factor of 0.500m K₂SO₄ will be highest.

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Van't Hoff factor was introduced for better understanding of colligative property of a solution.

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a) For NaCl the van't Hoff factor is 2

b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]

Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.

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3 years ago
A gas mixture is made up of kr (21.7 g), o2 (7.18 g), and co2 (14.8 g). the mixture has a volume of 23.1 l at 59 °c. calculate t
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Answer:- partial pressure of Kr = 0.306 atm, partial pressure of oxygen = 0.264 atm and partial pressure of carbon dioxide = 0.396 atm

Total pressure is 0.966 atm

Solution:- moles of Kr = 21.7 g x (1mol/83.8g) = 0.259 mol

moles of oxygen = 7.18 g x (1mol/32g) = 0.224 mol

moles of carbon dioxide = 14.8 g x (1mol/44g) = 0.336 mol

Volume of container = 23.1 L and the temperature is 59 + 273 = 332 K

From ideal gas law equation, P = nRT/V

partial pressure of Kr = (0.259 x 0.0821 x 332).23.1 = 0.306 atm

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Total pressure of the gas mixture = 0.306 atm + 0.264 atm + 0.396 atm = 0.966 atm

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