A 3.214 g sample of magnesium reacts with 8.416 g of bromine.The only product is magnesium bromide.If 1.934 g of magnesium is left unreacted,how much magnesium bromide is formed?
A)1.280 g
B)5.202 g
C)9.696 g
D)7.136 g
E)3.268 g

Question

Quizplus · Accepted Answer

The balanced chemical equation is:Mg + Br2 → MgBr2The molar mass of Mg is 24.31 g/mol, and the molar mass of Br2 is 159.81 g/mol.The moles of Mg reacted are:3.214 g Mg / 24.31 g/mol = 0.1322 mol MgThe moles of Br2 reacted are:8.416 g Br2 / 159.81 g/mol = 0.0527 mol Br2The mole ratio of Mg to Br2 is 1:1, so the limiting reactant is Br2.The moles of MgBr2 formed are:0.0527 mol Br2 × 1 mol MgBr2 / 1 mol Br2 = 0.0527 mol MgBr2The mass of MgBr2 formed is:0.0527 mol MgBr2 × 184.11 g/mol = 9.696 g MgBr2Therefore, the correct answer is C) 9.696 g.

A 3214 G Sample of Magnesium Reacts with 8