Identify if vanillin or hydride, is the limiting reagent and calculate the theor
ID: 891374 • Letter: I
Question
Identify if vanillin or hydride, is the limiting reagent and calculate the theoretical yield of vanillyl alcohol in grams.Vanillin Information: MW = 152 g/mol MP = 81-83 Celcius BP = 170/15 mm Density = 1.056
Hydride (HCl) Information: Concentration = 5 M Volume used = 4.5 mL
Vanillyl Alcohol Information: MW = 154.17 g/mol MP = 113-115 Celcius Amount Produced = 0.3589 g Identify if vanillin or hydride, is the limiting reagent and calculate the theoretical yield of vanillyl alcohol in grams.
Vanillin Information: MW = 152 g/mol MP = 81-83 Celcius BP = 170/15 mm Density = 1.056
Hydride (HCl) Information: Concentration = 5 M Volume used = 4.5 mL
Vanillyl Alcohol Information: MW = 154.17 g/mol MP = 113-115 Celcius Amount Produced = 0.3589 g
Vanillin Information: MW = 152 g/mol MP = 81-83 Celcius BP = 170/15 mm Density = 1.056
Hydride (HCl) Information: Concentration = 5 M Volume used = 4.5 mL
Vanillyl Alcohol Information: MW = 154.17 g/mol MP = 113-115 Celcius Amount Produced = 0.3589 g
Explanation / Answer
HCl molarity = 5 M
volume = 4.5 ml
molarity = (mass / molar mass ) x 1000/V
5 = (mass /36.5 ) x 1000 / 4.5
mass = 0.82 g
vanilin + HCl --------------------> vinilyl alcohol
152 g 36.5g 154.17 g
0.82g
36.5 g HCl needs --------------- 154.17 g vanillyl alcohl
0.82 g HCl --------------- ??
yield of vanillyl alcohol = 0.82 x 154.17 / 36.5
= 3.46 g
theoritical yield of vanillyl alcohol = 3.46 g
note : here only mention mass of HCl. so it is the limiting reagent