In a combination reaction, 2.22 g of Magnesium is heated with 3.75 g of Nitrogen
a) which reactant is present in excess?
b) how many moles of product are formed?
c) after reaction, how many grams of each reactant and product are present?
a) Well, it seems like Magnesium and Nitrogen decided to have a sizzling hot date! But when it comes to reactions, one of them is bound to be the "excess" party-goer. To figure out who's the life of the reaction, we need to do a little math.
First, we calculate the number of moles of Magnesium using its molar mass. Magnesium has a molar mass of about 24.31 g/mol, so we divide the given mass (2.22 g) by the molar mass to find the moles.
Next, we do the same for Nitrogen. Nitrogen has a molar mass of about 14.01 g/mol, so we divide the given mass (3.75 g) by the molar mass to find the moles.
Now, whichever reactant has fewer moles will be the "excess" one. So, let's compare the moles of Magnesium and Nitrogen and see who's bringing too much enthusiasm to the reaction!
b) Now that we know who is the life of the reaction and who's a bit too enthusiastic, let's move on to the next question: how many moles of product are formed? In a combination reaction, Magnesium and Nitrogen combine to form Magnesium Nitride (Mg3N2). This means that 3 moles of Magnesium react with 1 mole of Nitrogen to produce 1 mole of Magnesium Nitride.
Since we've determined the limiting reactant in the previous question, we can figure out how many moles of product are formed by looking at the reactant ratio. So, let's get calculating!
c) Last but not least, it's time to determine the masses of each reactant and product after the reaction. Remember, we found out who was the excess party-goer in question (a). So, we can use that information to calculate the remaining mass of the excess reactant and the mass of the product formed.
Take the limiting reactant (the one NOT in excess) and multiply its moles with its molar mass to get the mass of the limiting reactant consumed. Subtracting this mass from the initial mass of the limiting reactant will give us the remaining mass after the reaction. As for the product mass, we can calculate it by multiplying the moles of product with its molar mass.
Now, grab your calculators and let's calculate the masses and see who's left standing after this reaction!
Remember, math can be fun, especially when you add a bit of chemistry to the mix!
To determine which reactant is present in excess, we can compare the moles of each reactant and see which one has a greater quantity relative to the stoichiometry of the balanced equation.
a) To find the number of moles of each reactant, we need to divide the given mass of each reactant by their respective molar masses.
The molar mass of Magnesium (Mg) is 24.31 g/mol, and the molar mass of Nitrogen (N2) is 28.02 g/mol.
Moles of Magnesium (Mg) = Mass of Magnesium (g) / Molar mass of Magnesium (g/mol)
Moles of Magnesium = 2.22 g / 24.31 g/mol = 0.091 mol
Moles of Nitrogen (N2) = Mass of Nitrogen (g) / Molar mass of Nitrogen (g/mol)
Moles of Nitrogen = 3.75 g / 28.02 g/mol = 0.134 mol
According to the balanced equation for the combination of Magnesium and Nitrogen to form Magnesium Nitride (Mg3N2), the molar ratio is 3:1. That means for every 3 moles of Magnesium, we need 1 mole of Nitrogen.
Since we have 0.091 moles of Magnesium and 0.134 moles of Nitrogen, the Magnesium is present in excess because there is more Magnesium than the stoichiometric ratio requires.
b) To determine the number of moles of product formed, we need to use the stoichiometry of the balanced equation. From the balanced equation, we know that 3 moles of Magnesium react with 1 mole of Nitrogen to produce 1 mole of Magnesium Nitride.
Since Magnesium is present in excess, we can use the moles of Nitrogen (0.134 mol) to calculate the moles of the product (Magnesium Nitride).
Moles of product = Moles of Nitrogen = 0.134 mol
c) To calculate the mass of each reactant and product after the reaction, we can use the moles of each substance and their respective molar masses.
Mass of Magnesium after reaction = Moles of Magnesium (0.091 mol) × Molar mass of Magnesium (24.31 g/mol)
Mass of Nitrogen after reaction = Moles of Nitrogen (0.134 mol) × Molar mass of Nitrogen (28.02 g/mol)
Mass of product (Magnesium Nitride) = Moles of product (0.134 mol) × Molar mass of Magnesium Nitride (100.93 g/mol)
You can now calculate the masses by substituting the respective values into the equations.
To determine the answers to these questions, we need to follow a step-by-step approach using stoichiometry, which involves converting grams to moles and vice versa.
Let's start by writing the balanced chemical equation for the reaction between magnesium (Mg) and nitrogen (N2):
3 Mg + N2 -> 2 Mg3N2
a) To determine which reactant is in excess, we need to compare the amount of each reactant used to the stoichiometric ratio in the balanced equation. First, let's calculate the moles of each reactant:
Moles of Mg = mass of Mg / molar mass of Mg
Moles of N2 = mass of N2 / molar mass of N2
The molar mass of magnesium (Mg) is approximately 24.31 g/mol, and the molar mass of nitrogen (N2) is approximately 28.02 g/mol. Plugging in the values:
Moles of Mg = 2.22 g / 24.31 g/mol
Moles of N2 = 3.75 g / 28.02 g/mol
Calculate the moles for each reactant:
Moles of Mg ≈ 0.091 mol
Moles of N2 ≈ 0.134 mol
To determine the limiting reactant, we compare their stoichiometric ratios in the balanced equation. From the equation, we can see that 3 moles of Mg react with 1 mole of N2 to form 2 moles of Mg3N2. Thus, the mole ratio of Mg:N2 is 3:1.
Looking at the moles we calculated above, we see that we have fewer moles of Mg than N2. Therefore, magnesium (Mg) is the limiting reactant, and nitrogen (N2) is in excess.
b) To calculate the moles of product formed, we use the balanced chemical equation and the limiting reactant:
Moles of Mg3N2 = Moles of Mg * (2 moles of Mg3N2 / 3 moles of Mg)
Moles of Mg3N2 ≈ 0.091 mol * (2 mol Mg3N2 / 3 mol Mg)
Moles of Mg3N2 ≈ 0.061 mol
c) Lastly, let's calculate the mass of each reactant and the product after the reaction:
Mass of reactants and product = Moles * molar mass
Mass of Mg = Moles of Mg * molar mass of Mg
Mass of N2 = Moles of N2 * molar mass of N2
Mass of Mg3N2 = Moles of Mg3N2 * molar mass of Mg3N2
Plugging in the values:
Mass of Mg = 0.091 mol * 24.31 g/mol
Mass of N2 = 0.134 mol * 28.02 g/mol
Mass of Mg3N2 = 0.061 mol * (2 * (24.31 g/mol) + 28.02 g/mol)
Solving these equations will give you the mass of each reactant and the product after the reaction.
2.22 g of Mg is 2.22/24.31 = 0.09 mole
3.75g of N is 3.75/14.01 = 0.27 mole
Now, what compound is formed? The ratio of elements will affect how much reagent is used.
If we assume the simplest Mg3N2 then 3 moles of Mg are used for each 2 moles of N.
So, .09 mole of Mg will react with 2/3 * .09 = .06 mole of N.
You can take it from here, I guess.