# Aluminum acetate reacts with sodium iodide. If starting with 6.19 milligrams of aluminum acetate, how many grams of sodium iodide are required ?

May 30, 2018

The mass of $\text{NaI}$ required is 0.0136 g.

#### Explanation:

You know that you will need a balanced chemical equation with masses, moles, and molar masses, etc.

Step 1. Assemble all the information in one place

${M}_{\textrm{r}} : \textcolor{w h i t e}{m m m l l} 204.11 \textcolor{w h i t e}{m m l l} 149.89$
$\textcolor{w h i t e}{m m m m} {\text{Al"("C"_2"H"_3"O"_2)_3 + "3NaI" → "AlI"_3 + "3NaC"_2"H"_3"O}}_{2}$
$m \text{/mg} : \textcolor{w h i t e}{m m} 6.19$

Step 2. Calculate the moles of "Al"("C"_2"H"_3"O"_2)_3

"Moles of Al"("C"_2"H"_3"O"_2)_3 = 6.19 color(red)(cancel(color(black)("mg Al"("C"_2"H"_3"O"_2)_3))) × ("1 mmol Al"("C"_2"H"_3"O"_2)_3)/(204.11 color(red)(cancel(color(black)("mg Al"("C"_2"H"_3"O"_2)_3)))) = "0.030 33 mmol Al"("C"_2"H"_3"O"_2)_3

Step 3 Calculate the moles of $\text{NaI}$

$\text{Moles of NaI" = "0.030 33" color(red)(cancel(color(black)("mmol Al"("C"_2"H"_3"O"_2)_3)))× ("3 mmol NaI")/(1 color(red)(cancel(color(black)("mmol Al"("C"_2"H"_3"O"_2)_3)))) = "0.090 98 mmol NaI}$

Step 3. Calculate the mass of $\text{NaI}$

$\text{Mass of NaI" = "0.090 98" color(red)(cancel(color(black)("mol NaI"))) × "149.89 mg NaI"/(1 color(red)(cancel(color(black)("NaI")))) = "13.6 mg NaI" = "0.0136 g NaI}$