When magnesium burns in air, it reacts with oxygen and nitrogen, forming magnesium oxide and magnesium nitride. If 20.0L of air is available for this reaction, what mass of each compound can be formed? (Remember air is about 80% Nitrogen and 20% Oxygen)
2Mg + O2 ==> 2MgO
3Mg + N2 ==> Mg3N2
20 mL air x 0.80 N2 = 16 mL N2
20 mL air x 0.20 O2 = 4 mL O2.
I assume we are dealing with STP conditions; therefore,
mols N2 = 16/22.4 = ? and convert that to mols Mg3N2 then to grams. g = mols x molar mass.
mols O2 = 4/22.4 = ? and convert to mols MgO, then to grams.
Total = g MgO + grams Mg3N2.
To determine the mass of each compound formed when magnesium burns in air, we need to consider the balanced chemical equation for the reaction:
2Mg + O2 + N2 -> 2MgO + Mg3N2
From the equation, we can see that 2 moles of magnesium reacts with 1 mole of oxygen and 1 mole of nitrogen to form 2 moles of magnesium oxide and 1 mole of magnesium nitride.
First, let's calculate the moles of oxygen and nitrogen present in 20.0L of air:
Moles of oxygen = (20.0L)(0.20)(1 mole/22.4L) = 0.1786 moles
Moles of nitrogen = (20.0L)(0.80)(1 mole/22.4L) = 0.7143 moles
Next, we need to determine the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product formed. In this case, we compare the number of moles of magnesium to the number of moles of oxygen and nitrogen.
The ratio of moles of magnesium to moles of oxygen is 2:1 (from the balanced equation). Since we have 2 moles of magnesium, we need twice as many moles of oxygen.
Moles of oxygen needed = 2(2 moles of magnesium) = 4 moles
Since we only have 0.1786 moles of oxygen, it is the limiting reactant.
Now, let's calculate the moles of products formed:
Moles of magnesium oxide = 2 moles of magnesium oxide / 4 moles of oxygen needed = 0.5 moles
Moles of magnesium nitride = 1 mole of magnesium nitride / 4 moles of oxygen needed = 0.25 moles
Finally, we can calculate the mass of each compound using their respective molar masses:
Mass of magnesium oxide = Moles of magnesium oxide × Molar mass of magnesium oxide
Mass of magnesium nitride = Moles of magnesium nitride × Molar mass of magnesium nitride
The molar masses are:
Magnesium oxide (MgO) = 24.31 g/mol + 16.00 g/mol = 40.31 g/mol
Magnesium nitride (Mg3N2) = (3 × 24.31 g/mol) + (2 × 14.01 g/mol) = 100.94 g/mol
Now we can perform the calculations:
Mass of magnesium oxide = 0.5 moles × 40.31 g/mol = 20.16 g
Mass of magnesium nitride = 0.25 moles × 100.94 g/mol = 25.235 g
Therefore, approximately 20.16 grams of magnesium oxide and 25.235 grams of magnesium nitride can be formed when 20.0 L of air is available for this reaction.
To find the mass of each compound formed when magnesium burns in air, we need to consider the balanced chemical equation for the reaction:
2Mg + O2 + N2 -> 2MgO + Mg3N2
From the balanced equation, we can see that for every 2 moles of magnesium, 2 moles of oxygen and 1 mole of nitrogen are consumed, resulting in the formation of 2 moles of magnesium oxide (MgO) and 1 mole of magnesium nitride (Mg3N2).
Step 1: Convert the given volume of air into moles of oxygen (O2) and nitrogen (N2).
Since air is about 80% nitrogen and 20% oxygen, we can calculate the moles of oxygen and nitrogen present in the given volume of air as follows:
Moles of Oxygen (O2) = Volume of air x Percentage of Oxygen / Molar volume of Oxygen
Moles of Nitrogen (N2) = Volume of air x Percentage of Nitrogen / Molar volume of Nitrogen
The molar volume at standard conditions (0°C and 1 atm) is 22.4 L/mol for both oxygen and nitrogen.
For oxygen:
Moles of Oxygen (O2) = 20.0 L x 0.20 / 22.4 L/mol = 0.1786 mol
For nitrogen:
Moles of Nitrogen (N2) = 20.0 L x 0.80 / 22.4 L/mol = 0.7143 mol
Step 2: Determine the limiting reactant.
To determine the limiting reactant, we need to compare the number of moles of magnesium available with the moles of oxygen and nitrogen required for the reaction.
Given that we have 2 moles of magnesium, we can compare it with the moles of oxygen and nitrogen required as follows:
Oxygen required = 2 moles of oxygen per 2 moles of magnesium = 2 moles
Nitrogen required = 1 mole of nitrogen per 2 moles of magnesium = 1 mole
Since the number of moles of oxygen (0.1786 mol) and nitrogen (0.7143 mol) are both greater than the required amounts, the limiting reactant is magnesium.
Step 3: Calculate the moles and masses of the products formed.
Since magnesium is the limiting reactant, we will use its moles to calculate the moles of the products formed.
Moles of Magnesium Oxide (MgO) = Moles of Magnesium (Mg) = 2 moles
Moles of Magnesium Nitride (Mg3N2) = 1/2 of Moles of Magnesium (Mg) = 1 mole
To calculate the mass of each compound, we need to multiply the moles of each compound by their respective molar masses.
The molar mass of magnesium oxide (MgO) is 40.31 g/mol.
The molar mass of magnesium nitride (Mg3N2) is 100.93 g/mol.
Mass of Magnesium Oxide (MgO) = Moles of Magnesium Oxide (MgO) x Molar mass of MgO
= 2 mol x 40.31 g/mol
= 80.62 g
Mass of Magnesium Nitride (Mg3N2) = Moles of Magnesium Nitride (Mg3N2) x Molar mass of Mg3N2
= 1 mol x 100.93 g/mol
= 100.93 g
Therefore, the mass of magnesium oxide formed will be 80.62 grams, and the mass of magnesium nitride formed will be 100.93 grams.