Question

1. How many moles are in an 84.2 g sample of aluminum sulfate? ​

Answer 1

Answer:

0.246 moles

Explanation:

Aluminum Sulfate is an ionic equation so we need to write the equation first with proper charges and subscripts:

• Aluminum (Al) has a charge of [3+]. Sulfate (SO4) has a charge of [2-]. The charges need to be swapped to match.
• The ionic equation therefore is: Al_2(SO4)_3. We need 2 Aluminum's to match with the charge of 3 sulfates.

Now that we know the ionic equation, we need to find the molar mass of Al_2(SO4)_3. There's 2 Aluminum's, each weighing 26.98 g/mol. There are 3 sulfates, each sulfate weighs 96.06 g/mol. We need to multiply the number of elements/compounds by their mass to get the total molar mass and then add them up (remember atomic mass = molar mass in g/mol):

• (2 Al * 26.98 g/mol) + (3 (SO4) * 96.06 g/mol) =  342.15 g/mol of Al_2(SO4)_3

We now know the molar mass of Al_2(SO4)_3 we need to find how many moles of it there are in an 84.2g sample. To do that, we need to do some dimensional analysis:

• 84.2g Al_2(SO4)_3 * ( 1 mol of Al_2(SO4)_3 / 342.15 g/mol of Al_2(SO4)_3 )

The grams of Al_2(SO4)_3 cancel so that we're left with moles of Al_2(SO4)_3 on top:

• 84.2 * ( 1 mol of Al_2(SO4)_3 / 342.15 ) = 0.246 mol of Al_2(SO4)_3

So in an 84.2g sample of Al_2(SO4)_3 (aka aluminum sulfate), there are 0.246 moles. This is correct because in one mole of Al_2(SO4)_3, there are 342.15g and the sample of 84.2 g of Al_2(SO4)_3 is way smaller than that, so our answer should be less than one mole, which it is.

I hope this helped!