If 50 g of Magnesium(Mg) is reacted with 30 g of Bromine(Br), what is the mass of the products?
a
30 g
b
20 g
c
50 g
d
80 g
To find the mass of the products, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed in a chemical reaction and determines the maximum amount of product that can be formed.
To find the limiting reactant, we compare the amount of each reactant to the stoichiometric ratio in the balanced chemical equation.
The balanced chemical equation for the reaction between magnesium (Mg) and bromine (Br) is:
Mg + Br2 -> MgBr2
According to the stoichiometry of the equation, the molar ratio between Mg and Br2 is 1:1. This means that for every 1 mole of Mg, we need 1 mole of Br2.
First, let's convert the given masses of Mg and Br into moles using their respective molar masses:
Molar mass of Mg = 24.31 g/mol
Molar mass of Br2 = 159.81 g/mol
Moles of Mg = (50 g / 24.31 g/mol) ≈ 2.06 mol
Moles of Br2 = (30 g / 159.81 g/mol) ≈ 0.19 mol
Based on the molar ratios in the balanced equation, we see that the amount of Br2 (0.19 mol) is less than the amount of Mg (2.06 mol). This means that Br2 is the limiting reactant.
Now, to calculate the mass of the products, we need to determine the moles of MgBr2 formed using the limiting reactant:
From the balanced equation, we see that 1 mole of Br2 produces 1 mole of MgBr2.
Therefore, the moles of MgBr2 formed is equal to the moles of Br2: 0.19 mol.
Finally, we calculate the mass of MgBr2 using its molar mass:
Molar mass of MgBr2 = 184.11 g/mol
Mass of MgBr2 = (0.19 mol) x (184.11 g/mol) ≈ 34.98 g
Therefore, the mass of the products is approximately 34.98 g.
Answer: The mass of the products is approximately 34.98 g.
To answer this question, we need to determine the balanced chemical equation for the reaction between magnesium (Mg) and bromine (Br).
The balanced chemical equation is:
2 Mg + Br2 -> 2 MgBr
From the balanced chemical equation, we can see that for every 2 moles of magnesium reacted, we produce 2 moles of magnesium bromide (MgBr).
To find the mass of the products, we need to calculate the number of moles of magnesium (Mg) and bromine (Br) that are reacting.
Given:
Mass of magnesium (Mg) = 50 g
Mass of bromine (Br) = 30 g
To calculate the number of moles, we use the formula:
Number of moles = mass / molar mass
The molar mass of magnesium (Mg) is approximately 24.31 g/mol
The molar mass of bromine (Br) is approximately 79.90 g/mol
Number of moles of magnesium (Mg) = 50 g / 24.31 g/mol ≈ 2.06 mol
Number of moles of bromine (Br) = 30 g / 79.90 g/mol ≈ 0.38 mol
According to the balanced chemical equation, the ratio of magnesium (Mg) to magnesium bromide (MgBr) is 1:1. This means that for every 2 moles of magnesium (Mg) reacted, we produce 2 moles of magnesium bromide (MgBr).
Since we have 2.06 moles of magnesium (Mg), we can expect to produce 2.06 moles of magnesium bromide (MgBr) as well.
The molar mass of magnesium bromide (MgBr) is approximately 184.11 g/mol.
Mass of magnesium bromide (MgBr) = number of moles × molar mass
= 2.06 mol × 184.11 g/mol
≈ 378.77 g
Therefore, the mass of the products (magnesium bromide) is approximately 378.77 g.
The correct answer is not listed among the provided options.
To find the mass of the products, you need to determine the balanced chemical equation for the reaction between Magnesium (Mg) and Bromine (Br). The balanced equation for this reaction is:
2 Mg + Br₂ → 2 MgBr
From the balanced equation, you can see that 2 moles of magnesium react with 1 mole of bromine to produce 2 moles of magnesium bromide.
Now, let's calculate the number of moles of magnesium and bromine in the given quantities.
The molar mass of magnesium (Mg) is 24.31 g/mol, and 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
According to the balanced equation, 2 moles of magnesium react with 1 mole of bromine. Therefore, the limiting reactant is bromine because there is less of it. This means that only 0.38 moles of bromine react.
Using stoichiometry, we can determine the amount of magnesium bromide (MgBr₂) produced. From the balanced equation, 2 moles of magnesium bromide are produced for every 1 mole of bromine that reacts.
1 mol Br₂ → 2 mol MgBr₂
0.38 mol Br × (2 mol MgBr₂ / 1 mol Br₂) = 0.76 mol MgBr₂
Finally, let's calculate the mass of magnesium bromide (MgBr₂) produced using its molar mass, which is 184.12 g/mol.
0.76 mol MgBr₂ × (184.12 g MgBr₂ / 1 mol MgBr₂) ≈ 139.85 g
Therefore, the approximate mass of the products (magnesium bromide) formed in this reaction is 139.85 g.