Question

If 1.0 mol of Mg is mixed with 2.0 mol of Br2, and 0.88 mol of MgBr2 is obtained, what is the percent yield for the reaction?

Answer 1

Explanation:

Data given:

number of moles of magnesium = 1 mole

number of moles of bromine gas = 2 moles

moles of magnesium bromide obtained 0.42 moles

percent yield = ?

balance chemical reaction:

Mg Br2 -> MgBr2

1 mole of Mg reacted to give 1 mole of Mg Br2

1 mole of Br2 reacted to give 1 moles of Mg Br2

2 moles of Br2 reacted to give 2 moles of Mg Br2

limiting reagent is Mg so, mass obtained by 1 mole Mg 1 x184.11

mass obtained 184.11 (theoretical yield)

actual yield 0.42 x 184.11

actual yield 77.32 grams

percent yield

(actual yield/ theoretical yield) x 100

putting the values in the equation:

(77.32/184.11) x 100

percent yield 41.99 %