So I saw this question: If 28.0 grams of Pb(NO3)2 react with 18.0 grams of NaI, what mass of PbI2 can be produced? Pb(NO3)2 + NaI → PbI2 + NaNO3 I saw an answer that question but I need one part explained. The answer balanced the equation, but I do not know how they did that or why. They added a 2 to both sides and I do not know where it came from or why it made it balanced.

mass of PbI₂ = 27.6606 g

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

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