Question

what is the percent yield if you start with 25g C3H6 are produced when 33.5 g are used ?

C3H6 +Br2 >>C3H6Br2

Answer 1

formula

% yield = actual * 100 / theoretical mass

reaction

C₃H₆ + Br₂ -------> C₃H₆Br₂

molar mass of C₃H6 MM = 3 * MM(C) + 6 * MM(H)

= 3 * 12.01 + 6* 1.008

= 42.078 g/mol

mass (C₃H₆) = 25g

number of moles of (C₃H₆) n = mass of (C₃H₆) / molar mass of(C₃H₆)

n = 25g / 42.078 g/mol

n = 0.594 mol

molar mass of C₃H₆Br₂

MM = 3 * MM(C) + 6 * MM(H) + 2* MM(Br)

= 3 * 12.01 + 6* 1.008 + 2* 79.904

= 201.89 g/mol

according to the balanced equation

C₃H₆ + Br₂ -------> C₃H₆Br₂

mol of C₃H₆Br₂ formed = moles of C₃H₆

= 0.594 mol

use mass of CO₂ = number of mol* molar mass

= 0.594 * 201.89

= 119.92 g

% yield = actual * 100 / theoretical mass

= 33.5 * 100 / 119.92

= 27.9%

% yield = 27.9%

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