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?

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 +  ⇒ Mg

1 mole of Mg reacted to give 1 mole of Mg

1 mole of Br2 reacted to give 1 moles of Mg

2 moles of Br2 reacted to give 2 moles of Mg

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 =

x 100

putting the values in the equation:

x100

percent yield = 41.99 %

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