Solid magnesium metal reacts with chlorine gas according to the following equation:
Mg + Cl2 --> MgCl2
What mass of magnesium chloride is formed in the reaction between 7 g of Mg and 13.2 g of Cl2
This is a limiting reagent problem. You know that when BOTH reactants are given.
Convert 7 g Mg to moles. moles = grams/molar mass
Convert 13.2 grams Cl2 to moles.
Using the coefficients in the balanced equation, convert mole Mg to moles MgCl2.
Same process, convert moles Cl2 to moles MgCl2.
The answers to moles MgCl2 will be different; obviously, one of them is wrong. In limiting reagent problems, the correct answer is ALWAYS the smaller one and the reactant producing that value is the limiting reagent.
Now use the smaller value and convert moles MgCl2 to grams. g = moles x molar mass.
Post your work if you get stuck.
To determine the mass of magnesium chloride formed, we need to first find the limiting reagent. The limiting reagent is the reactant that is completely consumed and determines the amount of product that can be formed.
1. Calculate the number of moles of each reactant:
- Moles of Mg = mass / molar mass = 7 g / 24.31 g/mol = 0.288 mol
- Moles of Cl2 = mass / molar mass = 13.2 g / 70.90 g/mol = 0.186 mol
2. Determine the stoichiometry of the reaction from the balanced equation:
1 mol of Mg reacts with 1 mol of Cl2 to form 1 mol of MgCl2.
3. Compare the mole ratios of the reactants:
The mole ratio of Mg to Cl2 is 0.288 mol / 0.186 mol ≈ 1.55
Since the mole ratio is approximately 1.55, we can see that there is an excess of Mg compared to Cl2. Therefore, Cl2 is the limiting reagent.
4. Calculate the moles of MgCl2 formed:
Since the stoichiometry of the balanced equation tells us that 1 mol of MgCl2 is formed for every 1 mol of Cl2, the moles of MgCl2 formed is also 0.186 mol.
5. Calculate the mass of MgCl2 formed:
Mass of MgCl2 = moles of MgCl2 × molar mass of MgCl2
Mass of MgCl2 = 0.186 mol × (24.31 g/mol + 2 × 35.45 g/mol)
Mass of MgCl2 = 0.186 mol × 95.21 g/mol
Mass of MgCl2 = 17.68 g
Therefore, the mass of magnesium chloride formed in the reaction between 7 g of Mg and 13.2 g of Cl2 is 17.68 g.
To find the mass of magnesium chloride formed in the reaction between 7 g of Mg and 13.2 g of Cl2, we need to determine the limiting reactant first.
1. Calculate the number of moles for each reactant:
- Moles of Mg = mass of Mg / molar mass of Mg
Molar mass of Mg = 24.31 g/mol
Moles of Mg = 7 g / 24.31 g/mol
- Moles of Cl2 = mass of Cl2 / molar mass of Cl2
Molar mass of Cl2 = 35.45 g/mol
Moles of Cl2 = 13.2 g / 35.45 g/mol
2. Determine the stoichiometric ratio between Mg and MgCl2 using the balanced equation:
From the balanced equation, we can see that 1 mole of Mg reacts with 1 mole of Cl2 to produce 1 mole of MgCl2.
3. Identify the limiting reactant:
The reactant that provides the lowest number of moles is the limiting reactant. In this case, we compare the moles of Mg and Cl2 to find the limiting reactant.
Moles of MgCl2 = Moles of limiting reactant
Based on the stoichiometry, the moles of MgCl2 are equal to the number of moles of the limiting reactant.
4. Calculate the moles of MgCl2 using the limiting reactant:
Moles of MgCl2 = Moles of limiting reactant
Moles of MgCl2 = Moles of Mg (since Mg is the limiting reactant in this case)
5. Calculate the mass of MgCl2 formed:
Mass of MgCl2 = Moles of MgCl2 * molar mass of MgCl2
Molar mass of MgCl2 = 95.21 g/mol (24.31 g/mol for Mg + 70.90 g/mol for Cl2)
Mass of MgCl2 = Moles of Mg * molar mass of MgCl2
Now, let's substitute the values into the equation:
Moles of MgCl2 = Moles of Mg = 7 g / 24.31 g/mol
Moles of MgCl2 = 0.288 mol
Mass of MgCl2 = Moles of MgCl2 * molar mass of MgCl2
Mass of MgCl2 = 0.288 mol * 95.21 g/mol
Mass of MgCl2 = 27.41 g
Therefore, the mass of magnesium chloride formed in the reaction between 7 g of Mg and 13.2 g of Cl2 is 27.41 grams.