Question

Please calculate the % yield

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

10ml of 3-methyl-2-butanol

25ml of (12M HCl)

1.263g of 2-chloro-2-methylbutane were produced.

Answer 1

frist convert the mL of 3-methyl-2-butanol in to grams using

density = mass / volume formula

density of 3-methyl-2-butanol is 0.818 g/mL which is taken from wiki

mass of 3-methyl-2-butanol = density x volume = 0.818 g/moL x 10 mL = 8.18 g

moles of 3-methyl-2-butanol = mass / molar mass = 8.18 g / 88.15 g/mol = 0.0928 mol

moles of product = 1.263 g / 106.6 g/mol = 0.01185 mol

now

% of yield (moles of product / moles of reactant ) x 100

= (0.01185 / 0.0928) x 100

= 12.7 %