What is the maximum amount of magnesium chloride you can produce when 15 g of magnesium (Mg) reacts with 20 g of hydrochloric acid (HCl)?
Mg + HCl → MgCl2 + H2
To find the maximum amount of magnesium chloride (MgCl2) that can be produced, we need to determine which reactant is the limiting reagent. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
First, we need to calculate the number of moles of each reactant.
The molar mass of magnesium (Mg) is 24.31 g/mol. To calculate the number of moles of magnesium, we divide the mass (15 g) by the molar mass:
Number of moles of Mg = 15 g / 24.31 g/mol = 0.617 mol
The molar mass of hydrochloric acid (HCl) is 36.46 g/mol. To calculate the number of moles of hydrochloric acid, we divide the mass (20 g) by the molar mass:
Number of moles of HCl = 20 g / 36.46 g/mol = 0.548 mol
Now, we need to compare the moles of magnesium (Mg) and hydrochloric acid (HCl) to determine the limiting reagent.
From the balanced chemical equation, we can see that the ratio of moles of Mg to moles of HCl is 1:2. This means that for every mole of Mg, we need 2 moles of HCl.
Since the ratio of moles is 1:2, we can determine that the amount of hydrochloric acid (HCl) is limiting in this reaction. The reaction will completely consume 0.548 mol of HCl.
Now, let's calculate the maximum amount of magnesium chloride (MgCl2) that can be produced. From the balanced chemical equation, we can see that the ratio of moles of MgCl2 to moles of HCl is also 1:2. This means that for every mole of HCl, we can produce 1 mole of MgCl2.
Since we have 0.548 mol of HCl, we will produce an equal amount (0.548 mol) of MgCl2.
Finally, to calculate the mass of magnesium chloride, we multiply the number of moles by the molar mass. The molar mass of MgCl2 is 95.21 g/mol.
Maximum amount of MgCl2 = 0.548 mol × 95.21 g/mol = 52.14 g
Therefore, the maximum amount of magnesium chloride that can be produced is 52.14 grams.
To determine the maximum amount of magnesium chloride (MgCl2) that can be produced when 15 g of magnesium (Mg) reacts with 20 g of hydrochloric acid (HCl), we need to follow these steps:
1. Convert the masses of Mg and HCl into moles using their respective molar masses.
2. Determine the limiting reactant by comparing the number of moles of each compound.
3. Use the stoichiometry of the balanced equation to find the number of moles of MgCl2 produced.
4. Convert the number of moles of MgCl2 into grams using its molar mass.
Let's calculate it step-by-step:
Step 1: Convert the masses of Mg and HCl into moles.
- Molar mass of Mg = 24.31 g/mol
- Molar mass of HCl = 36.46 g/mol
Moles of Mg = Mass of Mg / Molar mass of Mg
Moles of Mg = 15 g / 24.31 g/mol
Moles of HCl = Mass of HCl / Molar mass of HCl
Moles of HCl = 20 g / 36.46 g/mol
Step 2: Determine the limiting reactant.
To find the limiting reactant, compare the moles of each reactant.
The balanced equation tells us that the stoichiometric ratio between Mg and HCl is 1:2.
Since the coefficient of Mg is 1 and the coefficient of HCl is 2, the number of moles of HCl is twice the number of moles of Mg.
Therefore, we need to multiply the moles of Mg by 2 to see if it is equal to or less than the moles of HCl.
Moles of Mg = (15 g / 24.31 g/mol) * 2 = 1.96 mol
Moles of HCl = 20 g / 36.46 g/mol = 0.55 mol
Since 1.96 mol of Mg is greater than 0.55 mol of HCl, HCl is the limiting reactant.
Step 3: Use the stoichiometry of the balanced equation to find the moles of MgCl2.
From the balanced equation, the stoichiometric ratio between HCl and MgCl2 is 2:1.
This means that 2 moles of HCl react to produce 1 mole of MgCl2.
Since we have 0.55 mol of HCl (the limiting reactant), we can calculate the moles of MgCl2 produced.
Moles of MgCl2 = 0.55 mol * (1 mol MgCl2 / 2 mol HCl) = 0.275 mol MgCl2
Step 4: Convert the moles of MgCl2 into grams.
To convert moles of MgCl2 to grams, we need to multiply by the molar mass of MgCl2.
Molar mass of MgCl2 = atomic mass of Mg + 2*(atomic mass of Cl)
Molar mass of MgCl2 = (24.31 g/mol) + 2*(35.45 g/mol) = 95.21 g/mol
Mass of MgCl2 = Moles of MgCl2 * Molar mass of MgCl2
Mass of MgCl2 = 0.275 mol * 95.21 g/mol = 26.17 g
Therefore, the maximum amount of magnesium chloride that can be produced when 15 g of Mg reacts with 20 g of HCl is approximately 26.17 grams.
This is a limiting reagent (LR) problem.
Write and balance the equation. You have that but it isn't balanced.
Mg + 2HCl ==> MgCl2 + H2
Convert what you have top mols
mols Mg = 15/24.3 = approx 0.6 but you will need to ALL of these calculations over since I've estimated all of them.
mols HCl = 20/36.5 = about 0.5
Convert mols of each reactant to mols of product using the coefficients in the balanced equation.
0.6 mols Mg x (1 mol MgCl2/1 mol Mg) = about 0.6
0.5 mols HCl x (1 mol MgCl2/2 mol HCl) = about 0.25.
In LR problems the SMALLER number always wins so HCl is the LR.
Using the LR, calculate grams product formed.
about 0.25 mols x molar mass MgCl2 = grams MgCl2 produced.
Post your work if you get stuck.