Determine the theoretical yield in grams and the percent yield of the reaction if you start with 3.00 moles of H2 and 2.00 moles of I2 to produce 1.00 moles of HI.

Answer 1

Equation:

H₂ + I₂ ---> 2HI

Theoretical yield (T_y):

3.00 mol H₂ x (2 mol HI / 1 mol H₂) = 6 mol HI

2.00 mol I₂ x (2 mol HI / 1 mol I₂) = 4 mol HI

4 mol HI x (127.911g / mol HI) = 511.644g HI

Percent yield (P_y):

1 mol HI x (127.911g / mol HI) = 127.911g HI

4 mol HI x (127.911g / mol HI) = 511.644g HI

P_y = (A_y / T_y) x 100%

P_y = (127.911g / 511.644g) x 100%

P_y = 0.25 (100%)

P_y = 25%