Question

question 5
1 pts
according to the balanced chemical equation from the previous question, if 1.5 g of lead (ii) nitrate are reacted with 2.5 g of potassium iodide, what is the limiting reactant?
lead (ii) nitrate
potassium iodide

Explanation:

Step1: Calculate moles of lead (II) nitrate

The molar mass of \(Pb(NO_3)_2\) is \(M = 207.2+(2\times14)+(6\times16)=331.2\space g/mol\).
The number of moles \(n_{Pb(NO_3)_2}=\frac{m}{M}=\frac{1.5\space g}{331.2\space g/mol}\approx0.00453\space mol\)

Step2: Calculate moles of potassium iodide

The molar mass of \(KI\) is \(M = 39.1 + 126.9=166\space g/mol\).
The number of moles \(n_{KI}=\frac{m}{M}=\frac{2.5\space g}{166\space g/mol}\approx0.0151\space mol\)

Assume the balanced chemical equation is \(Pb(NO_3)_2 + 2KI
ightarrow PbI_2+2KNO_3\)

From the stoichiometry, the mole ratio of \(Pb(NO_3)_2:KI = 1:2\)

For \(n_{Pb(NO_3)_2} = 0.00453\space mol\), the moles of \(KI\) required \(n_{KI,required}=2\times0.00453 = 0.00906\space mol\)

Since \(n_{KI,available}=0.0151\space mol>n_{KI,required} = 0.00906\space mol\)

For \(n_{KI} = 0.0151\space mol\), the moles of \(Pb(NO_3)_2\) required \(n_{Pb(NO_3)_2,required}=\frac{0.0151}{2}=0.00755\space mol\)

Since \(n_{Pb(NO_3)_2,available}=0.00453\space mol < n_{Pb(NO_3)_2,required}=0.00755\space mol\)

Answer:

Lead (II) nitrate