Question

5. How many grams of AlCl3 are produced when 0.5 mL of a 2.5 M Al2(SO4)3 solution are used in the reaction?

6 NaCl + Al2(SO4)3 -----> 3 Na2SO4 + 2 AlCl3

Answer 1

we have the Balanced chemical equation as:

6 NaCl + Al2(SO4)3 -----> 3 Na2SO4 + 2 AlCl3

lets calculate the mol of Al2(SO4)3

volume , V = 0.5 mL

= 5*10^-4 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 2.5*0.0005

= 1.25*10^-3 mol

From balanced chemical reaction, we see that

when 1 mol of Al2(SO4)3 reacts, 2 mol of AlCl3 forms

mol of AlCl3 formed = (2/1)* moles of Al2(SO4)3

= (2/1)*1.25*10^-3

= 2.5*10^-3 mol

This is number of moles of AlCl3

Molar mass of AlCl3 = 1*MM(Al) + 3*MM(Cl)

= 1*26.98 + 3*35.45

= 133.33 g/mol

we have below equation to be used:

mass of AlCl3,

m = number of mol * molar mass

= 2.5*10^-3 mol * 133.33 g/mol

= 0.3333 g

Answer: 0.333 g