"Silver nitrate, AgNO3, reacts with ferric chloride, FeCl3, to give silver chloride, AgCl, and ferric nitrate, Fe(NO3)3. In a particular experiment, it was planned to mix a solution containing 25.0 g of AgNO3 with another solution containing 45.0 grams of FeCl3. 3 AgNO3 + 1 FeCl3 ---> 3 AgCl + 1 Fe(NO3)3

What is the maximum number of grams of AgCl or also known as the "theoretical yield" that could be obtained from these starting quantities?"

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g