What's the empirical formula of 36.48% na 25.41% s 38.11% o

A compound is found to contain 36.48% Na, 25.41% S, and 38.11% O. find its empirical formula. % in such problems are by weight, & can be changed to grams out of a 100 grams total using molar mass, find moles: 
36.48g Na @ 23 g/mol = 1.6 moles
Na 25.41g S, @ 32 g/mole = 0.8 mol
S 38.11g O. @ 16 g/mol = 2.4 mol
O ratio the moles, by dividing all by the smallest: 1.6 mol Na / 0.8 = 2 mol Na 0.8 mol S / 0.8 = 1 mole
 S 2.4 mol O / 0.8 = 3 mol O 

your empirical formula os Na2SO3 

It was hard when I took chemistry too! I passed with a B

The activity series is shown in the attached picture. if the element is below the other, then you can displace the other. following the activity series as a guide, the reactions are as follows: nickel and aluminum chloride3 ni + 2 alcl₃  → 3 nicl₂ + 2 albarium and copper bromideba + cubr  →  no reaction because cu is less oxidizable than bapotassium nitrate and aluminum3 kno₃ + al  → al(no₃)₃ + 3 ksodium phosphate and zincnaso₄ + zn  → znso₄ + na