Question

A chemistry student reacts 2 g of lead (Ii) nitrate with 4 g of potassium iodide. how much product in grams should the student expect?

Answer 1

The mass of the product, PbI₂ (s), the student should expect is 2.784 grams.

Step-by-step explanation

Given:

Mass of lead (II) nitrate = 2 g

Mass of potassium iodide = 4 g

Using the atomic masses elements in the periodic table;

Molar mass of lead (II) nitrate (Pb(NO₃)₂) = 331.2 g/mol

potassium iodide (KI) = 166.0028 g/mol

What to find:

The product in grams that is produced.

Step-by-step solution:

Step 1: Write a balanced chemical equation for the reaction.

Pb(NO₃)₂ (aq) + 2KI (aq) → 2KNO₃ (aq) + PbI₂ (s)

Step 2: Convert the given masses to moles

Using the mole formula

`\begin{gathered} Moles=\frac{Mass}{Molar\text{ }mass} \\ \\ Moles\text{ }of\text{ }Pb\left(NO₃\right)₂=\frac{2\text{ }g}{331.2\text{ }g\text{/}mol}=0.006039\text{ }mol \\ \\ Moles\text{ }of\text{ }KI=\frac{4\text{ }g}{166.0028\text{ }g\text{/}mol}=0.024096\text{ }mol \end{gathered}`

Step 3: Determine the limiting reactant.

There are more moles of potassium iodide than of lead (II) nitrate (from step 2), but the ratio is only the following:

`\frac{0.024096\text{ }mol\text{ }KI}{0.006039\text{ }mol\text{ }Pb\left(NO₃\right)₂}=3.99`

Because the ratio of the coefficients in the balanced chemical equation is,

`\frac{2\text{ }mol\text{ }KI}{1\text{ }mol\text{ }Pb\left(NO₃\right)₂}=2`

There is not enough lead (II) nitrate (Pb(NO₃)₂) to react with all the potassium iodide (KI), thus, lead (II) nitrate (Pb(NO₃)₂) is the limiting reactant that determines the grams of the product produced.

Step 4: Calculate the moles of the product ( PbI₂) produced using the mole ratio in step 1 and the number of moles of Pb(NO₃)₂ in step 2.

`\begin{gathered} 1\text{ }mol\text{ }KI=1\text{ }mol\text{ }PbI_2 \\ \\ 0.006039\text{ }mol\text{ }KI=x \\ \\ x=\frac{0.006039\text{ }mol\text{ }KI}{1\text{ }mol\text{ }KI}*1\text{ }mol\text{ }PbI_2=0.006039\text{ }mol\text{ }PbI_2 \end{gathered}`

Step 5: Convert 0.006039 mol PbI₂ to grams using the same formula in step 2.

`\begin{gathered} Mass=Moles* Molar\text{ }mass \\ \\ Substitute\text{ }moles=0.006039\text{ }mol\text{ }and\text{ }molar\text{ }mass\text{ }of\text{ }PbI_2=461.01\text{ }g\text{/}mol \\ \\ Mass=0.006039\text{ }mol*461.01\text{ }g\text{/}mol \\ \\ Mass=2.784\text{ }grams \end{gathered}`

Therefore the mass of the product, PbI₂ (s), the student should expect is 2.784 grams.