When an aqueous solution of lead(ii) nitrate is mixed with an aqueous solution of sodium iodide, an aqueous solution of sodium nitrate and a yellow solid, lead iodide, are formed?

When an aqueous solution of lead(ii) nitrate (Pb(NO₃)₂) is mixed with an aqueous solution of sodium iodide (NaI), an aqueous solution of sodium nitrate (NaNO₃) and a yellow solid, lead iodide (PbI₂), are formed.

The balanced equation for the reaction is:

 Pb(NO₃)₂(aq)+2NaI(aq)→2NaNO₃(aq)+PbI₂(s)

lithosphere and asthenosphere

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