Suppose you tried to combine 42 grams of magnesium with 42 grams of oxygen.
Which of the two substances would have some left after the reaction and how much magnesium oxide would be formed?
When one pound of gasoline (made up of compounds of hydrogen and carbon) is burned in an automobile approximately 3 pounds of carbon dioxide, CO2, is given off. Carbon dioxide is one of the gases contributing to global warming. What information from this experiment helps explain how one pound of gasoline can give off approximately 3 times an much CO2?
To determine which substance would have some left after the reaction, we need to calculate the stoichiometry of magnesium and oxygen in the reaction.
The balanced equation for the reaction between magnesium and oxygen is:
2Mg + O2 → 2MgO
From this equation, we can see that 2 moles of magnesium react with 1 mole of oxygen to form 2 moles of magnesium oxide.
To calculate the amount of magnesium oxide formed, we need to compare the number of moles of each reactant.
1. Calculate the number of moles of magnesium:
Given mass of magnesium = 42 grams
Molar mass of magnesium (Mg) = 24.31 grams per mole
Number of moles = given mass / molar mass
Number of moles of magnesium = 42 g / 24.31 g/mol = 1.73 moles
2. Calculate the number of moles of oxygen:
Given mass of oxygen = 42 grams
Molar mass of oxygen (O2) = 32.00 grams per mole
Number of moles of oxygen = given mass / molar mass
Number of moles of oxygen = 42 g / 32.00 g/mol = 1.31 moles
Based on the stoichiometry of the balanced equation, you can see that 1.31 moles of oxygen are required to react with 2 moles of magnesium. Since we only have 1.31 moles of oxygen available, magnesium is the limiting reactant in this case.
Therefore, magnesium will be completely consumed in the reaction, and there will be some oxygen left over.
To calculate the amount of magnesium oxide formed, we can use the stoichiometry of the balanced equation. Since 2 moles of magnesium produce 2 moles of magnesium oxide, we can conclude that 1.73 moles of magnesium will produce 1.73 moles of magnesium oxide.
Therefore, 1.73 moles of magnesium oxide will be formed.
To determine which substance would have some left after the reaction and how much magnesium oxide would be formed, we need to consider their respective molar masses and the stoichiometry of the reaction.
First, let's find the molar masses of magnesium (Mg) and oxygen (O₂):
- Mg: The atomic mass of Mg is 24.31 g/mol.
- O₂: The molecular weight of O₂ (two oxygen atoms) is 32.00 g/mol.
Next, we need to find the number of moles of each substance by dividing the given mass by their respective molar masses:
- Moles of Mg: 42 g / 24.31 g/mol ≈ 1.729 mol
- Moles of O₂: 42 g / 32.00 g/mol ≈ 1.313 mol
Now, let's write the balanced chemical equation for the reaction between magnesium and oxygen:
2 Mg + O₂ → 2 MgO
From the balanced equation, we can see that for every two moles of magnesium, one mole of oxygen is required to form two moles of magnesium oxide.
Now, we need to determine the limiting reactant, which is the reactant that is completely consumed during the chemical reaction and limits the amount of product formed. To do this, we compare the number of moles of each reactant to the stoichiometric ratio:
The stoichiometric ratio between Mg and O₂ is 2:1.
Since the ratio of moles of Mg to moles of O₂ is approximately 1.729:1.313, magnesium (Mg) is the limiting reactant.
Using the stoichiometry, we can determine the amount of magnesium oxide (MgO) formed based on the limiting reactant:
- Moles of MgO formed = Moles of Mg (limiting reactant) ≈ 1.729 mol
Lastly, let's calculate the mass of magnesium oxide formed by multiplying the moles of MgO by its molar mass:
- Mass of MgO formed = Moles of MgO formed × Molar mass of MgO
The molar mass of MgO can be calculated by adding the atomic masses of magnesium (Mg: 24.31 g/mol) and oxygen (O: 16.00 g/mol):
- Molar mass of MgO = 24.31 g/mol + 16.00 g/mol = 40.31 g/mol
Now, substituting the values, we can calculate the mass of magnesium oxide formed:
- Mass of MgO formed = 1.729 mol × 40.31 g/mol ≈ 69.72 g
Therefore, after the reaction is complete, all 42 grams of magnesium (Mg) would be consumed, and approximately 69.72 grams of magnesium oxide (MgO) would be formed. There would be no magnesium or oxygen left over.
Limiting reagent problems are done alike.
1. Write and balance the equation.
2Mg + O2 ==> 2MgO
2. Convert grams to moles.
a. 42 grams oxygen/32 = 1.31 moles O2.
b. 42 grams Mg/24.305 = 1.73 moles Mg.
3. Using the coefficients in the balanced equation, convert moles in 2a and 2b to moles of the product.
a. from O2: moles MgO = 1.31 moles O2 x (2 moles MgO/1 mole O2) = 1.31 x 2 = 2.62 moles MgO formed.
b. from Mg: moles MgO = 1.73 moles Mg x (2 moles MgO/2 moles Mg) = 1.73 x 2/2 = 1.73 moles MgO formed.
c. USUALLY, moles of the product from a and from b are different, as they are in this case. You know both answers can't be right. Which one is right? The smaller value is ALWAYS the correct value AND THE MATERIAL PRODUCING THAT NUMBER IS THE LIMITING REAGENT (in this case Mg).
4.a. If the problem asks (this one does in the last part) for grams MgO formed, that's just the smaller value x molar mass; i.e., 1.73 moles MgO x molar mass MgO = grams MgO.
b. If the problem asks which reactant is left over, you have that information in step 3. The limiting reagent is Mg; therefore, oxygen must be the material left.
c. Some of these problems (this one doesn't) asks for how much of the "other" material is left over? You do that exactly as in step 3.
moles O2 used = 1.73 moles Mg x (1 mole O2/2 mols Mg) = 1.73/2 = 0.865 mols O2 used. We had 2.62 initially - 0.865 moles used = 1.75 moles O2 remaining un-reacted. You can conert that to grams by 1.75 g O2 x molar mass O2 = ??
You need to go through this again and assign good numbers to each part. I've rounded here and there and estimated at some other spots. Put this procedure somewhere so you can refer to it later.