Question

ed chemical equation. However, the balanced chemical equation only works in units of moles. To convert grams of aluminum to moles of aluminum the atomic mass of aluminum must be used. grams of aluminum x 1 mole Al = 1 26.98 g Al To relate the moles of aluminum to KAl(SO4)2* 12H20, the mole to mole ratio must be found in the balanced chemical equation. In this case it is: 1 mole of Al3 = 1 mole KAl(SO4)2 12H20 As a conversion factor it can be written in one of two ways: 1 mole KAJ(SO4)2 12H2O 1 mole of Al*3 1 mole of Al3 1 mole KAI(SO4)2 or 12H20 Use dimensional analysis to cancel units of moles of aluminum to moles of KAI(SO4)2 12H2O. 12H20 1 mole Al x 1 mole KAl(SO42 1 mole of Alt3 12H20 = 1 mole of KAl(SO4)2 1 Finally, moles of KAl(SO4)2 12H20 can be converted to grams of KAl(SO4)2 by calculating the formula mass. 12H20 1 mole KAl(SO42 12H20 x Formula mass KA|(SO42 12H O= theoretical yield 1 mole KAl(SO4)2 12H20 1 Don't forget to include the mass of the water in the formula mass.

Answer 1

ANSWER:-

3)-atomic mass of aluminium is 26.98 g/mole

4)- moles of aluminium = w/m = 1.01 / 26.98 = 0.038 moles

5)- moles of KAl (SO₄)₂ .12H₂O = W/M MASS OF KAl (SO₄)₂ .12H₂O is 474.37 gm/mole

moles of KAl (SO₄)₂ .12H₂O = 16.87/474.37 = 0.0355moles

6)- formula MASS OF KAl (SO₄)₂ .12H₂O is 474.37 gm/mole

7)- theoretical mass of KAl (SO₄)₂ .12H₂O is

by the formula

0.0355 mole of KAl (SO₄)₂ .12H₂O/1 x 474.37 formula mass of KAl (SO₄)₂ .12H₂O /1 mole of KAl (SO₄)₂ .12H₂O

= 16.84 gm

8)- percent yield = actual yield / theoretical yield x 100

percent yield = 16.87 / 16.84 x 100

= 100.17 %