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

1 Answer
May 30, 2018

Answer:

The mass of #"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_text(r):color(white)(mmmll)204.11color(white)(mmll)149.89#
#color(white)(mmmm)"Al"("C"_2"H"_3"O"_2)_3 + "3NaI" → "AlI"_3 + "3NaC"_2"H"_3"O"_2#
#m"/mg": color(white)(mm)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 #"NaI"#

#"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 #"NaI"#

#"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"#