If 50 g of Magnesium(Mg) is reacted with 30 g of Bromine(Br), what is the mass of the products?
a
50 g
b
80 g
c
30 g
d
20 g
To determine the mass of the products, we need to determine the limiting reactant in the reaction and use the stoichiometry of the balanced equation.
First, let's write the balanced equation for the reaction between magnesium and bromine:
2 Mg + Br2 → 2 MgBr
The molar mass of magnesium (Mg) is 24 g/mol and the molar mass of bromine (Br2) is 159.8 g/mol.
Using the molar masses, we can calculate the number of moles for each reactant:
- Moles of Mg = 50 g / 24 g/mol = 2.08 mol
- Moles of Br2 = 30 g / 159.8 g/mol = 0.19 mol
According to the balanced equation, the stoichiometric ratio between Mg and Br2 is 2:1. However, we have less than 1 mole of Br2, so it is the limiting reactant. This means that all of the bromine will be consumed and the amount of magnesium will be in excess.
Using the stoichiometry of the balanced equation, we can calculate the mass of the products:
- Moles of MgBr = 0.19 mol of Br2 * (2 mol of MgBr / 1 mol of Br2) = 0.38 mol
- Mass of MgBr = 0.38 mol * (24 + 79.9) g/mol = 39.902 g
Therefore, the mass of the products is approximately 39.902 g.
None of the given answer choices match this value exactly.