The balanced chemical equation for the reaction between magnesium and bromine is:
2 Mg + Br2 -> 2 MgBr
From the equation, we can see that 2 moles of magnesium react with 1 mole of bromine to produce 2 moles of magnesium bromide.
To find the moles of magnesium and bromine, we can use their respective molar masses:
Molar mass of Mg = 24.3 g/mol
Molar mass of Br = 79.9 g/mol
Moles of Mg = Mass of Mg / Molar mass of Mg
Moles of Mg = 50 g / 24.3 g/mol = 2.06 mol
Moles of Br = Mass of Br / Molar mass of Br
Moles of Br = 30 g / 79.9 g/mol = 0.38 mol
Since the reaction ratio is 2:1 for magnesium to bromine, we can see that bromine is the limiting reactant here. This means that all the bromine will react, and the amount of magnesium bromide produced will be determined by the amount of bromine.
From the balanced equation, we know that 1 mole of bromine produces 2 moles of magnesium bromide. Therefore, the moles of magnesium bromide produced will be:
Moles of MgBr = Moles of Br x (2 moles of MgBr / 1 mole of Br)
Moles of MgBr = 0.38 mol x (2 mol/1 mol) = 0.76 mol
Finally, to find the mass of the products (magnesium bromide):
Mass of MgBr = Moles of MgBr x Molar mass of MgBr
Mass of MgBr = 0.76 mol x (24.3 g/mol + 79.9 g/mol)
Mass of MgBr = 0.76 mol x 104.2 g/mol ≈ 79.2 g
Therefore, the mass of the products (magnesium bromide) is approximately 79.2 g.
The correct answer is (d) 80 g.
If 50 g of Magnesium(Mg) is reacted with 30 g of Bromine(Br), what is the mass of the products?
a
20 g
b
30 g
c
50 g
d
80 g
1 answer