How many grams of aluminum sulfate are produced from 2.67 mol sulfuric acid With the balanced equation Al2O3 3H2SO4 -- Al2 (SO4)3 3H2O?

Answer 1

Al2O3 + 3H2SO4 = Al2 (SO4)3 + 3H2O

We note that the molar ratios are

1:3::1:3

By Equivalence

0.89 : 2.67 ;; 0,89 :2.67

So moles of aluminium sulphate produced is 0.89

Using the equation

moles = mass(g) / Mr

Then mass = moles X Mr

Refer to the Atomic Masses in the Periodic Table to find the Mr(Relative Molecular mass of Al2(SO4)3

Hence

Al x 2 = 27 x 2 = 54

S x 3 = 32 x 3 = 96

O x (4 x 3 = 12) 16 x 12 = 192

54 + 96 + 192 = 342 ( Mr of Al2(SO4)3 )

Hence

mass(g) = 0,89 X 342 = 304.38 g (This is the 100% theoretical yield. Experimentally it will be less , owing to losses in the filter paper , spillages etc., )

Answer 2

To find the grams of aluminum sulfate produced from 2.67 mol of sulfuric acid, you must use the stoichiometry of the equation. Since the molar ratio is 3:1 for sulfuric acid to aluminum sulfate, you would multiply 2.67 mol by the stoichiometric ratio (1 mol / 3 mol) to find the mol of aluminum sulfate produced. Then you can convert the mol of aluminum sulfate to grams by using the molar mass of Al2(SO4)3.