How many aluminium atoms are present in 5.396 grams of aluminium? 2Al + 6HCI -> 2AlCl3 + 3H2

1 Answer
May 13, 2017

Answer:

Approx. #1.2xx10^23# individual aluminum atoms.

Explanation:

We need (i) to calculate the molar quantity:

#(5.396*g)/(26.982*g*mol^-1)=0.2000*mol#

And then (ii), multiply this molar quantity by #"Avogadro's number"#, #6.022xx10^23*mol^-1# to get...................................

#0.2*molxx6.022xx10^23=??#