we wish to determine the moles of pbi₂ precipitated when 125 ml of 0.20 m ki reacts with excess pb(no₃)₂…

we wish to determine the moles of pbi₂ precipitated when 125 ml of 0.20 m ki reacts with excess pb(no₃)₂. 2ki(aq) + pb(no₃)₂(aq) → 2kno₃(aq) + pbi₂(s) in the previous step, you determined 0.025 mol ki react. how many moles of pbi₂ form during the reaction?

we wish to determine the moles of pbi₂ precipitated when 125 ml of 0.20 m ki reacts with excess pb(no₃)₂. 2ki(aq) + pb(no₃)₂(aq) → 2kno₃(aq) + pbi₂(s) in the previous step, you determined 0.025 mol ki react. how many moles of pbi₂ form during the reaction?

Answer

Explanation:

Step1: Identify mole ratio

From the reaction ( 2\text{KI}(aq) + \text{Pb(NO}_3\text{)}_2(aq) \rightarrow 2\text{KNO}_3(aq) + \text{PbI}_2(s) ), the mole ratio of ( \text{KI} ) to ( \text{PbI}_2 ) is ( 2:1 ).

Step2: Calculate moles of ( \text{PbI}_2 )

Given moles of ( \text{KI} = 0.025 , \text{mol} ). Let ( n ) be moles of ( \text{PbI}2 ). Using the ratio ( \frac{n{\text{PbI}2}}{n{\text{KI}}} = \frac{1}{2} ), so ( n_{\text{PbI}_2} = \frac{0.025}{2} = 0.0125 , \text{mol} ).

Answer:

( 0.0125 , \text{mol} )