WHOEVER ANSWERS WILL GET BRAINLIEST Calculate the percentage of yield when 20 grams of sodium chloride solution reacts with an excess amount of silver nitrate solution knowing that 45 grams of silver chloride precipitated

%yield = 91.8

Given

20 g NaCl

45 g AgCl

Required

%yield

Solution

Reaction

NaCl + AgNO₃ ⇒ AgCl + NaNO₃

mol NaCl :

= mass : MW

= 20 g : 58,44 g/mol

= 0.342

mol AgCl from equation :

= 1/1 x mol NaCl

= 1/1 x 0.342

= 0.342

Mass AgCl(theoretical) :

= mol x MW

= 0.342  x 143,32 g/mol

= 49.02 g

%yield = (actual/theoretical) x 100%

%yield = (45/49.02) x 100%

%yield = 91.8

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8