If 1.50g lead(II) nitrate is reacted with 1.75g sodium chromate what is the theoretical yield of the precipitate?

1.46g of PbCrO₄ are the theoretical yield

Explanation:

Theoretical yield is defined as the maximum amount of products that could be produced (Assuming a yield of 100%).

The reaction of Lead (II) nitrate with sodium chromate is:

Pb(NO₃)₂(aq) + Na₂CrO₄(aq) → PbCrO₄(s) + 2NaNO₃ (aq)

First, we need to find molar mass of each reactant in order to determine limiting reactant (As the reaction is 1:1, the reactant with the lower number of moles is the limiting reactant). The moles of the limiting reactant = moles of Lead (II) chromate (The precipitate):

Moles Pb(NO₃)₂ -Molar mass: 331.21g/mol-

1.50g * (1mol / 331.21g) = 4.53x10⁻³ moles Pb(NO₃)₂

Moles Na₂CrO₄ -Molar mass: 161.98g/mol-

1.75g * (1mol / 161.98g) = 0.0108 moles

Pb(NO₃)₂ is limiting reactant and moles of PbCrO₄ are 4.53x10⁻³ moles. The mass is:

4.53x10⁻³ moles PbCrO₄ * (323.19g / mol) =

the answer to your question is c

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