Question

What is the percent yield of hbr if 85.00 g of hbr was formed from 30. g of h2?

Answer 1

1) Chemical equation

H2 (g) + Br2 (g) ---> 2 HBr

2) molar ratios

1 mol H2 : 1 mol Br2 : 2 moles HBr

3) convert 30.0 g of H2 to number of moles

moles = mass in grams / molar mass = 30.0 g / 2.02 g/mol = 14.85 moles

4) calculate theoretical yield

2 moles HBr / 1 mol H2 = x / 14.85 moles H2

=> x = 14.85 moles H2 * 2 moles HBr / 1 mol H2 = 29.7 moles HBr

5) convert 59.52 moles HBr to mass

mass = number of moles * molar mass = 29.7 moles * 80.91 g /mol = 2403.03 g

6) percent yield = (actual yield / theoretical yield) * 100 = (85.00g / 2403.03g)*100 = 3.54%

Answer: 3.54%

Answer 2

Ok, lets see the definitions of percent yield, actual yield and theoretical yield.

Percent yield is the ratio of actual yield to theoretical yield.

Actual yield- amount of product produced in the EXPERIMENT.

Theoretical yield- max amount of product produced through CALCULATIONS

% yield= actual yield (from experiment)/theoretical yield (from calculation) *100

1st Step: Write the reaction 

H2 + Br2 --> 2 HBr

2nd Step: Get the mass ratio of H2 and HBr to find theoretical yield

1 mole H2 gives 2 moles HBr ( molar mass of H2= 2 g/mol, HBr= 81 g/mol)

2 g H2 gives 2*81= 162 g HBr

30 g H2 gives 162*30/2 = 2430 g HBr ( The equation is 2 g H2/ 30 gH2= 162 g HBr/ x g HBr)

So theoretically 2430 g HBr are produced by calculation ( THEORETICAL YIELD)

By experiment 85 g HBr are produced as it is given at the question ( ACTUAL YIELD)

% yield= actual yield / theoretical yield *100 = 85/2430 *100= 3.5 %

The percent yield is 3.5 %.