Question

Need to calculate yeild by limiting reagent.

Reagents used:

Started with 2-methyl-2-butanol = 9.312g

HCL: 25ml

NaCl: 15 ml

Product:

2-chloro-2-methylbutane: 4.302g

Just attaching the experiment for reference

Answer 1

2-methyl-2-butanol + HCl ---------------> 2-chloro-2-methylbutane + H2O

Moles of 2-methyl-2-butanol = mass/ molar mass = 9.312 g/ 88 g/mol = 0.1058 mol

Moles of HCl:

volume of HCl = 25 ml

mass of HCl = density x volume = 0.00149 g/ml x 25 ml = 0.03725 g

moles of HCl = mass/ molar mass= 0.03725 g/ 36.5 g/mol = 1.02 x 10^-3 mol = 0.00102 mol

Hence,

moles of HCl are less than moles of 2-methyl-2-butanol .

Therefore,

HCl is the limiting reagent.

Then, yield is calculated based on limiting reagent HCl.

Yield:

2-methyl-2-butanol + HCl ---------------> 2-chloro-2-methylbutane + H2O

1 mol 1 mol

0.00102 mol ? = 0.00102 mol

Hence,

mass of  2-chloro-2-methylbutane = moles x molar mass= 0.00102 mol x 106.5 g/ mol = 0.108 g

But,

2-chloro-2-methylbutane obtained = 4.302g

So, HCl is not the limiting reagent.

----------------------------------------------------------------------------------------------------------------------------

2-methyl-2-butanol is the limiting reagent.

Yield is calculated based on 2-methyl-2-butanol.

2-methyl-2-butanol + HCl ---------------> 2-chloro-2-methylbutane + H2O

1 mol 1mol

0.1058 mol ? = 0.1058 mol

Hence,

mass of  2-chloro-2-methylbutane = moles x molar mass= 0.1058 mol x 106.5 g/ mol = 11.27 g

This is called theoretical yield of the product.

But,

2-chloro-2-methylbutane obtained = 4.302g

This is actual yield .

Hence,

Percent yield of the product = ( actual yield/ theoretical yield) x100

= 4.302/  11.27 g x 100

= 38.17 %