Help with this IB chemistry question please?

0.502 g of an alkali metal sulfate is dissolved in water and excess barium chloride solution,
BaCl aq) 2(aq), is added to precipitate all the sulfate ions as barium sulfate, BaSO4 (s). The
precipitate is filtered and dried and weighs 0.672 g.

Determine the molar mass of the alkali metal sulfate and state its units. [2 marks]

The full question:
(a) Calculate the amount (in mol) of barium sulfate formed. [2]
(b) Determine the amount (in mol) of the alkali metal sulfate present. [1]
(c) Determine the molar mass of the alkali metal sulfate and state its units. [2]
(d) Deduce the identity of the alkali metal, showing your working. [2]
(e) Write an equation for the precipitation reaction, including state symbols. [2]

X is an alkali metal (Period 1)
(SO4)2- is an anion with charge -2
thus the balanced compound is X2SO4
2X + (SO4)2- --> X2SO4

mass of X2SO4 = 0.502 g

a) moles of BaSO4?
Precipitate Mass = 0.672 g
M = 233.4 g mol-1
n = 0.00288 mol

b) X2SO4 + BaCL2 --> BaSO4 + 2 XCL
n(X2SO4)=n(BaSO4)
n=0.00288 mol

c) Molar mass of X2SO4?
m/n = M
0.502 g/0.00288 mol = 174.3 g/mol

d) 2 M(X) + 1 M(S) + 4 M(O) = 174.3 g/mol
2 M(X) = 78.3 g/mol
M(X) = 39.15 g/mol
the alkali metal is Potassium/ K (M=39.10 g/mol)

e) K2SO4 (aq) + BaCl2 (aq) --> BaSO4 (s) + 2 KCl (aq)

What Exam is this question from?

How is the number of moles of the alkali metal sulfate equals to that of BaSO4 dont get it