Question

If 1.0 mol of Mg is mixed with 2.0 mol of Br2, and 0.42 mol of MgBr2 is obtained, what is the percent yield for the reaction?

Answer 1

The percent yield of the reaction is 41.99 % If 1.0 mol of Mg is mixed with 2.0 mol of Br2, and 0.42 mol of MgBr2 is obtained.

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 +$Br_{2}$  ⇒ Mg$Br_{2}$

1 mole of Mg reacted to give 1 mole of Mg$Br_{2}$

1 mole of Br2 reacted to give 1 moles of Mg$Br_{2}$

2 moles of Br2 reacted to give 2 moles of Mg$Br_{2}$

limiting reagent is Mg

so, mass obtained by 1 mole Mg = 1 x 184.11

mass obtained = 184.11 (theoretical yield)

actual yield = 0.42 x 184.11

actual yield = 77.32 grams

percent yield =

$\frac{actual yield}{theoretical yield}$ x 100

putting the values in the equation:

$\frac{77.32}{184.11}$ x100

percent yield = 41.99 %