# What molar quantity of PbI_2 can be formed from a 5.85*mol quantity of "potassium iodide"?

May 1, 2017

Approx. $2.93 \cdot m o l$ $P b {I}_{2}$.

#### Explanation:

We need to interrogate the stoichiometric reaction:

$P {b}^{2 +} + 2 {I}^{-} \rightarrow P b {I}_{2} \left(s\right) \downarrow$

If there are $5.85 \cdot m o l$ $K I$ then at most we can AT most get $2.925 \cdot m o l$ $P b {I}_{2}$.

SO 2 questions:

what mass of $P b {I}_{2}$ does this molar quantity represent;

what is the colour of $P b {I}_{2}$?