How many grams of kbr are present in 250.0ml of a 1.20m solution

Molar mass KBr = 119 g/mol

Volume in liters: 250.0mL / 1000 => 0.25 L

n = M x V

n = 1.20 x 0,25 => 0.3 moles of KBr

Therefore:

1 mole KBr 119 g
0.3 moles KBr ??

0.3 x 119 / 1 => 35.7 g of KBr 

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