What mass of silver nitrate (agno3) is required to prepare 400 g of a 3.50% solution of agno3?

mass % = mass of the solute/   mass of the solvent

let mass of the solute= y

mass of solvent= 400g

%mass= 3.50 %  into fraction= 3.5/100

=3.5/100= y/400

multiply both side by 400

y= 14 grams

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