Question

Search PREPARATION OF ALUMLast saved by user Last Modified: Sat at BUM Design Layout References Mailings Review View Help ody) -14 A A A A E 21 ABCDd AaBbCcDd AaBbc AaBbcc AaB AaBbCcDAoBbccd U XX A. D A E B 1 Normal "No Spac... Heading 1 Heading 2 Title Subtitle Subtle Em... (aa) + AP (a + 2004 (aa) +12 MUM) 7 INAU4 12 1272010) Styles Note: Aluminium metal is the Limiting reagent. PERCENTAGE YIELD CALCULATION: Percentage yield = 100 x Actual yield, /Theoretical yield Molar mass of Alum = 474.08g/mol For every mole of Aluminum used = One mole of Alum was produced. Moles of alum theoretical = moles of aluminum used mass of alum expected mass of aluminium molar mass of alum molar mass of aluminium Actual yield= Measured mass from experiment Theoretical yield = mass of alum expected. Percentage yield is between 0% - 100%

Answer 1

Moles of Al used = mass/molar mass = 0.788g/27g/mol = 0.0292 mol of Aluminum.

Theoretical moles of alum = moles of Aluminum used = 0.0292 moles alum .

Theoretical mass of alum = theoretical moles × molar mass = 0.0292moles × 474.08g/mol = 13.843g alum .

% yield = (experimental yield/theoretical yield)×100

= (28.406g/13.843g)×100 = 205.2 %

This %yield (above 100%) is not possible at all , your alum product must be wet at weighing time .