Answer to Question #335585 in Inorganic Chemistry for Gf dsfa

Question

if 15.50 grams of lead (II) nitrate mixes with 3.81 grams of sodium chloride how many grams of sodium nitrate are formed

Answer 1

Pb(NO₃)₂ + 2NaCl = 2NaNO₃ + PbCl₂

n(Pb(NO₃)₂) = 15.50 g / 331.2 g/mol = 0.047 moles

n(NaCl) = 3.81g / 58.44 g/mol = 0.065 moles

NaCl - limiting reagent

m( NaNO₃₎ = (m(NaCl)*2Mr(NaCl)) / 2Mr(NaNO₃)

m( NaNO₃) = ( 3.81 g * (2*58.44 g/mol)) / 2*84.99 g/mol = 2.62 g