HINDI KO SURE TO. Cinopy ko lang sa binigay na lab rep nung upperclass :) 

4.) A compound of the element A and oxygen of has a mole ratio A: O =2:3. If 8.0 grams of the oxide contains 2.4 grams of the oxygen,

a.) What is the atomic weight of A?

b.) What is the weight of one mole of the oxide?

c.) What theoretical weight of the oxide is formed when 28 grams of A is heated in excess oxygen? What is the % yield, if 38 grams of the oxide was produced?

a.) 8.0 - 5.6 = 2.4 g O
2.4 g O x 1 mole O / 16 g O = 0.15 moles O

A:O = 2:3
A:0.15 = 2:3
{A}/{0.15} x {2}/{3}
3A = (0.15)(2)
3A = 0.30
A = 0.1 moles
AW = g/mole
AW = 5.6 g O / 0.1 moles O
AW = 56 g/mole

b)
2:3
A2O3
= (56)(2) + (16)(3)
= 160 g oxide

c)    *theoretical weight of oxide
A + O2                 A2O3
4A + 3O2                 2A2O3

1 mole A x 2moles A2O3 / 4moles A = 0.5
.5 moles A x 2 moles A2O3 / 4moles A = 0.25
LR: 0.25
= 0.25 x 160
= 40 g

* percent yield

%yield= (actual)/(theoretical) x 100

 = (38)over(40)  x 100

% = 95

ml-mmol

Volume A

1.      2ml x 0.25mole/ml= 0.5mol

2.      4ml x 0.25mole/ml= 1 mol

3.      6ml x 0.25mole/ml= 1.5 mol

4.      8ml x 0.25mole/ml= 2 mol

5.      10ml x 0.25mole/ml= 2.5 mol

Excess reagent:

1.       1mol -0.5mol= 0.5

2.       1mol – 1mol= 0.0

3.       1.5mol – 1mol= 0.5

4.       2mol – 1mol=  1

5.       2.5mol -mol1= 1.5

HINDI KO SURE TO  4.) A compound of the element A and oxygen of has a mole ratio A: O =2:3. If 8.0 grams of the oxide contains 2.4 grams of the oxygen, a.) What is the atomic weight of A? b.) What is the weight of one mole of the oxide? c.) What theoretical weight of the oxide is formed when 28 grams of A is heated in excess oxygen? What is the % yield, if 38 grams of the oxide was produced? a.) 8.0 - 5.6 = 2.4 g O 2.4 g O x 1 mole O / 16 g O = 0.15 moles O A:O = 2:3 A:0.15 = 2:3 {A}/{0.15} x {2}/{3} 3A = (0.15)(2) 3A = 0.30 A = 0.1 moles AW = g/mole AW = 5.6 g O / 0.1 moles O AW = 56 g/mole b) 2:3 A2O3 = (56)(2) + (16)(3) = 160 g oxide c) *theoretical weight of oxide A + O2 A2O3 4A + 3O2 2A2O3 1 mole A x 2moles A2O3 / 4moles A = 0.5 .5 moles A x 2 moles A2O3 / 4moles A = 0.25 LR: 0.25 = 0.25 x 160 = 40 g * percent yield t= (actual)/(theoretical) x 100 = (38)over(40) x 100 % = 95