calculate the no. of aluminium ions in 0.051 g of aluminium oxide.?

Question

hint: the mass of an ion is same as that of an atom of the same element.
al=27amu

Answer 1

Greetings!

Molecular Weight (Al2O3)= 2x27 + 3x16 = 102 g/mol
0.051g (Al2O3) / 102 g/mol = 0.0005 mol
1 (Al2O3) = 2Al(+) + 3 O(-)

Al(+) = 2x0.0005 = 0.001mol!!
O(-) = 3x0.0005 = 0.0015mol!!!

Good luck

Answer 2

In 102 g of Aluminium oxide we have 54 grams of aluminium (2 moles)
hence in 0.051 g we have 0.027g(0.001 moles)
1 mole of Al has 6.023 * 10^23 atoms hence 0.001 mole has 6.023*10^20 atoms

Answer 3

oxygen content 3x16 =48
alluninium = 2x27=54
total =54+48=102
no of alluminium atoms = (.051/102)x(2) 6.023x10(pw 23)
= 6.023x10(pwr 20) atoms

Answer 4

0.051gAl2O3* (1molAl2O3/101.964gAl2O3) *(2molAl/1molAl2O3) *(6.02*10^23atoms/1molAl)= 6.02212546 × 10^20 Al ions. The units cancel out, leaving out Al ions.

Answer 5

at mass of Al=26.98 at mass of O=16 TOTAL mol.mass of Al2o3=2*26.98+3*16.
no. of moles =.051/(molecular mass of Al2O3)
no. of ions=no.of moles* avogadro no.(i.e.6.023*10^23)

Answer 6

calculate as : Al2O3 total 102g mol

6.023x10^23 mol 6.23x10^23 ------ 102g
X ----------------- 0,051 g
then it is : 0,051 x 6.23 x10^23
-------------------------- =
102 g