Given 5.0 grams of lead (II) nitrate and 3.0
grams of sodium iodide, how many grams of sodium nitrate can be formed?

                     1.70 g of NaNO₃

Explanation:

                     The balance chemical equation for given double displacement reaction is,

                               Pb(NO₃)₂ + 2 NaI → PbI₂ + 2 NaNO₃

Step 1: Calculate moles of each reactant:

                                   Moles =  Mass / M.Mass

For Pb(NO₃)₂:

                                   Moles =  5.0 g / 331.21 g/mol

                                   Moles =  0.0150 moles

For NaI:

                                   Moles =  3.0 g / 149.89 g/mol

                                   Moles =  0.020 moles

Step 2: Calculate Limiting reagent as;

According to equation,

                     1 mole of Pb(NO₃)₂ reacts with  =  2 moles of NaI

So,

              0.0150 moles of Pb(NO₃)₂ will react with  =  X moles of NaI

Solving for X,

                     X =  2 mol × 0.0150 mol / 1 mol

                     X =  0.030 moles of NaI

Hence, it means that NaI is the limiting reagent therefore, it will control the yield of Sodium Nitrate.

Step 3: Find out moles of NaNO₃ formed:

According to equation,

                     2 mole of NaI produced =  2 moles of NaNO₃

So,

              0.020 moles of NaI will produce   =  X moles of NaNO₃

Solving for X,

                     X =  2 mol × 0.020 mol / 2 mol

                     X =  0.020 moles of NaNO₃

Step 4: Calculate Mass of NaNO₃ produced;

As,                Moles  =  Mass / M.Mass

Or,

                    Mass  =  Moles × M.Mass

Putting values,

                    Mass  =  0.020 mol × 84.99 g/mol

                    Mass  =  1.70 g of NaNO₃

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your answer would be a. 12.5 kg