26g --- 1 mol
56g --- X
X= 56/26 = 2,154 mol
959 ml = 959cm³ = 0,959dm³
C = n/V
C = 2,154/0,959
C = 2,246 mol/dm³
mass of PbI₂ = 27.6606 g
<h3>Further explanation</h3>
Given
Pb(NO₃)₂ + NaI → PbI₂ + NaNO₃
28.0 grams of Pb(NO₃)₂ react with 18.0 grams of NaI
Required
mass of PbI₂
Solution
Balanced equation
Pb(NO₃)₂ + 2NaI → PbI₂ + 2NaNO₃
The principle of a balanced reaction is the number of atoms in the reactants = the number of atoms in the product
mol Pb(NO₃)₂ :
= 28 : 331,2 g/mol
= 0.0845
mol NaI :
= 18 : 149,89 g/mol
= 0.12
Limiting reactant : mol : coefficient
Pb(NO₃)₂ : 0.0845 : 1 = 0.0845
NaI : 0.12 : 2 = 0.06
NaI limiting reactant (smaller ratio)
mol PbI₂ based on NaI
= 1/2 x 0.12 = 0.06
Mass PbI₂ :
= 0.06 x 461,01 g/mol
= 27.6606 g
