Respuesta :
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!
