A 3.214-g sample of magnesium reacts with 8.416g of bromine. The only product is magnesium bromide. If 1.934g of magnesium is left unreacted, how much magnesium bromide is formed?

Question

-Sorry, I've completely forgot how to do this..please help

Answer 1

Write the equation and balance it.
Mg + Br2 ==> MgBr2
Convert 3.214 g Mg to moles. moles = grams/molar mass
Convert 8.416 g bromine to moles using th same procedure.
Since you know bromine is the limiting reagent (since Mg is in excess), work this as in any stoichiometry problem.
Convert the 1.934 g Mg to moles and subtract from moles you had to start with. That will give you the moles Mg that reacted. Then convert that to moles MgBr2 using the coefficients in the balanced equation.
moles Mg used x (1 mole MgBr2/1 mole Mg) = moles Mg used x 1 = moles MgBr2 produced.
Then g MgBr2 = moles MgBr2 x molar mass MgBr2.