Question

How many atoms of each element are present in 3.2 moles of k2so3

Answer 1

Answer:- $3.85*10^2^4$ atoms of K, $1.93*10^2^4$ atoms of S and $5.78*10^2^4$ atoms of O.

Solution:- We have 3.2 moles of $K_2SO_3$ . Looking at it's molecular formula, it has two K, one S and three O. It means, 1 mole of the compound has two moles of K, one mole of S and 3 moles of O.

Now we could calculate the moles of each present in 3.2 moles of the compound as..

So, moles of K = 3.2 mol of compound x (2 mol K/mol of compound) = 6.4 mol K

moles of S = 3.2 mol of compound x (1mol S/1mol compound) = 3.2 mol S

moles of O = 3.2 mol of compound x (3 mol O/1mol compound) = 9.6 mol O

Now to convert the moles of each to atoms, we multiply by Avogadro number.

atoms of K = 6.4 mol x (6.022 x 10^23 atoms/1mol) = $3.85*10^2^4$ atoms of K

atoms of S = 3.2 mol x (6.022 x 10^23 atoms/1mol) = $1.93*10^2^4$ atoms of S

atoms of O = 9.6 mol x (6.022 x 10^23 atoms/1mol) = $5.78*10^2^4$ atoms of O

Answer 2

1.9264*10^23  is the answer