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
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.
To determine the mass of the products, we need to first write and balance the chemical equation for the reaction between Magnesium (Mg) and Bromine (Br). The balanced equation is:
2 Mg + Br2 -> 2 MgBr
From the balanced equation, we can see that 1 mole of Magnesium reacts with 1 mole of Bromine to produce 2 moles of Magnesium Bromide (MgBr2).
To find the moles of Magnesium and Bromine in the given masses, we need to divide the mass of each element by its molar mass.
The molar mass of Magnesium (Mg) is 24.31 g/mol.
The molar mass of Bromine (Br) is 79.90 g/mol.
For Magnesium:
50 g Mg * (1 mol Mg / 24.31 g Mg) = 2.06 mol Mg
For Bromine:
30 g Br * (1 mol Br / 79.90 g Br) = 0.38 mol Br
From the balanced equation, we know that Magnesium and Bromine react in a 1:1 ratio. However, Magnesium is in excess in this reaction since we have more moles of Magnesium (2.06 mol) compared to moles of Bromine (0.38 mol).
Therefore, Bromine will be fully consumed in the reaction, and we can calculate the moles of Magnesium Bromide (MgBr2) formed using the moles of Bromine.
0.38 mol Br * (2 mol MgBr2 / 1 mol Br) = 0.76 mol MgBr2
Finally, to find the mass of the products, we multiply the moles of Magnesium Bromide by its molar mass.
0.76 mol MgBr2 * (24.31 g MgBr2 / 1 mol MgBr2) = 18.54 g MgBr2
Therefore, the mass of the products formed in the reaction between 50 g of Magnesium and 30 g of Bromine is approximately 18.54 grams.
So, the correct choice is option a) 20 g.
To determine the mass of the products, we first need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction, thereby limiting the amount of product that can be formed.
To find the limiting reactant, we can compare the moles of each reactant using their respective molar masses.
The molar mass of Magnesium (Mg) is 24.31 g/mol.
The molar mass of Bromine (Br) is 79.90 g/mol.
First, we calculate the number of moles for each reactant:
Number of moles of Mg = mass of Mg / molar mass of Mg
Number of moles of Mg = 50 g / 24.31 g/mol
Number of moles of Mg ≈ 2.06 mol
Number of moles of Br = mass of Br / molar mass of Br
Number of moles of Br = 30 g / 79.90 g/mol
Number of moles of Br ≈ 0.38 mol
Now, we need to determine the reaction stoichiometry. The balanced equation for the reaction between Mg and Br is:
2Mg + Br2 -> 2MgBr
According to the balanced equation, the molar ratio between Mg and Br is 2:1. This means that for every 2 moles of Mg, we need 1 mole of Br to react completely.
Comparing the moles of Mg and Br, we see that there are more moles of Mg (2.06 mol) than moles of Br (0.38 mol). Therefore, Mg is the excess reactant and Br is the limiting reactant.
To find the mass of the products, we need to calculate the moles of MgBr2 formed:
Number of moles of MgBr2 = moles of Br x (2 moles of MgBr2 / 1 mole of Br)
Number of moles of MgBr2 = 0.38 mol x (2 mol MgBr2 / 1 mol Br)
Number of moles of MgBr2 = 0.76 mol
Next, we calculate the mass of MgBr2:
Mass of MgBr2 = number of moles of MgBr2 x molar mass of MgBr2
Mass of MgBr2 = 0.76 mol x (95.21 g/mol for MgBr2)
Mass of MgBr2 ≈ 72.37 g
Therefore, the mass of the products is approximately 72.37 g. Hence, the correct answer is option d) 80 g.