Question

Pb(NO₃)₂ + 2 NaI à PbI₂ + 2 NaNO₃

4)   If 100 grams of lead (II) nitrate and 100 grams of sodium iodide were initially used in the reaction, which is the limiting reactant? Which is the excess reactant?

 

 

5)   Find the amount excess in grams in #4

Answer 1

Molar mass of Lead (ii) nitrate $=331.2g/mol$

Molar mass of sodium Iodide $=149.89g/mol$

Moles of sodium iodide $=\frac{100}{149.89}=0.6672$

Moles of Lead (ii) Nitrate $=\frac{100}{331.2}=0.3019$

Reacting ratio $=1:2$

$0.3019×2=0.6038<0.6672$

Limiting reagent is Pb(NO₃)₂

Excess reagent is NaI