How many #Mg^(2+)# ions are present in 3.00 moles of #MgCl_2#?

1 Answer
Aug 6, 2018

Answer:

#1.81" x "10^24 Mg^(2+) "ions" #

Explanation:

Step 1:
From the formula #MgCl_2#, we know that it's made from 1 #Mg^(2+)# and 2 #Cl^-# ions

  • In 1 mol of #MgCl_2#, there's 1 mol of #Mg^(2+)#

We can use this relationship to calculate the moles of #Mg^(2+) "ions"# from the given 3.00 moles of #MgCl_2#.

Step 2:
Once we find the moles of #Mg^(2+)#, we can find the number of #Mg^(2+) "ions"# using Avogadro's number.

  • 1 mol #Mg^(2+)# has #6.022# x #10^23# #Mg^(2+) "ions"#

Combining the 2 steps calculations:
#Mg^(2+)"ions"="3.00 mols " MgCl_2 xx ("1 mol " Mg^(2+))/("1 mol " MgCl_2)xx(6.022 " x " 10^23 Mg^(2+) ions)/("1 mol " Mg^(2+))=1.81" x "10^24 Mg^(2+) "ions" #