Answer:
0.5352 grams ≈ 0.54 grams
Explanation:
The stoichiometry  of the equation is given below:
AgNO3 (aq) + NaCl (aq) ⇒ AgCl (s) + NaNO3 (aq)
1 mole       1 mole      1 mole     1 mole
The mole ratio between the reactants  AgNO3 and NaCl is 1 : 1
From the question, the  number of moles of AgNO3 used can be calculated using the formula below:
n = CV/1000
where n= number of moles
      C = concentration in mol/dm3
      V = Volume in mL
n (AgNO3) = 25 x 0.366/1000
         =  0.00915‬ moles
From the equation of reaction, I mole of AgNO3 reacts with 1 mole of NaCl; so 0.00915 moles of AgNO3 will react with exactly 0.00915 moles of NaCl.
To covert moles  of NaCl to grams, we use the formula below:
Number of moles (n) = Mass in grams (m)/ Molar mass (M)
n= m/M
n = 0.00915 moles
m =?
M = Molar mass of NaCl  [23 + 35.5= 58.5 grams]
m = n x M
  = 0.00915 x 58.5
  =  0.5353 grams
 ≈ 0.54 grams
