What molar quantity of $P b I_{2}$ $PbI_2$ can be formed from a $5.85 \cdot m o l$ $5.85*mol$ quantity of $potassium iodide$ $"potassium iodide"$ ?

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Answer 1 · 2017-05-01T13:27:09

Approx. $2.93 \cdot m o l$ $2.93*mol$ $P b I_{2}$ $PbI_2$ .

We need to interrogate the stoichiometric reaction:

$P b^{2 +} + 2 I^{-} \to P b I_{2} (s) ⏐ ↓$ $Pb^(2+) + 2I^(-) rarr PbI_2(s)darr$

If there are $5.85 \cdot m o l$ $5.85*mol$ $K I$ $KI$ then at most we can AT most get $2.925 \cdot m o l$ $2.925*mol$ $P b I_{2}$ $PbI_2$ .

SO 2 questions:

what mass of $P b I_{2}$ $PbI_2$ does this molar quantity represent;

what is the colour of $P b I_{2}$ $PbI_2$ ?

What molar quantity of PbI2PbI_2 can be formed from a 5.85⋅mol5.85*mol quantity of potassium iodide"potassium iodide"?