Question

Calculate the percent yield of your product assuming you started with 0.545 grams of Al metal and collect 5.25 grams of alum product?

1) Calculate the number of moles of Al used in your reaction.

2) Calculate the molar mass of alum, potassium aluminum sulfate dodecahydrate (do not forget to include the associated water in you molar mass calculation).

3) This experiment has been set up such that Al is the limiting reagent,(determines the maximum amount of product that could theoretically be formed) and other reagents are present in excess. Using the equation for the overall reaction in the lab instructions, one mole of Al can produce how many moles of alum?

4) Based on the number of moles of Al used in your reaction, calculate the maximum number of grams of alum (the theoretical yield of alum) that could be prepared. (start with grams Al and get to how many grams of Alum you would expect, theoretically).

5) Calculate the percent yield of alum in your reaction. % yield = (actual yield/ theoretical yield) x 100

Answer 1

Balanced Rxh is a Lal+ 2KOH + 4H2804 + 22H30 Ikalonte 1940]+3H 1. No of moles of Al= Mass - Molar Mass 0.545 g -0.0202mal 26.981591mal 2. Molar Mass of KAI(SO4)2 • 12 H₂O = Atomic mass of k . - + Atomic mass of Al +(Atomic mass of SX 2) + (Atomic mass of Ox 20)+ (Atomic mass of HX 24) =(39.0983) #26.9815) +(3.2.064 X2)+ (15.999X 20)+(1.60798201 = 474.39 g/mol According to balanced Rxh I moles Al give 2 moles Alum : 1 mole Al would give I mole Alum 3. A. No of holes of Alum formed = No of mules of Aluminium Consumed 0.0202 med 5. Theoretical Yield of Alum= No of moles x Molar Mass _= 0.0204 mol x_474.39 g/mol = 9.58279 Too olo Yield = obtained Yield x 100l Theoretical hield = 5.25g x10ot. 9.58279 = 54.8018 w