The moles of silver iodide produced are 5.00 moles.
Given:
The balanced chemical equation:
[tex]MgI (aq)+2AgNO_3 (aq)\rightarrow 2 Agl(s) + Mg(NO_3)_2(aq)[/tex]
To find:
The moles of silver iodide produced when 2.50 moles of magnesium iodide react with sufficient silver nitrate.
Solution:
Moles of magnesium iodide = 2.50 mol
[tex]MgI(aq) +2AgNO_3 (aq)\rightarrow 2 Agl(s) + Mg(NO_3)_2(aq)[/tex]
According to the reaction, 1 mole of magnesium iodide gives 2 moles of silver iodide, then 3.50 moles of magnesium iodide will give:
[tex]=\frac{2}{1}\times 2.50 mol=5.00 \text{mol of AgI}[/tex]
The moles of silver iodide produced are 5.00 moles.
Learn more about the unitary method here:
brainly.com/question/4147368?referrer=searchResults
brainly.com/question/2499283?referrer=searchResults
