When 16.3 g of magnesiumm reacts with 4.52 g of oxygen, how many grams of magnesium oxide will be performed? Identify the limiting and excess reactants.
To calculate the grams of magnesium oxide formed, we need to determine the limiting and excess reactants.
The balanced chemical equation for the reaction between magnesium and oxygen to form magnesium oxide is:
2Mg + O2 -> 2MgO
To find the limiting reactant, we compare the moles of each reactant.
First, calculate the moles of magnesium using the given mass:
Molar mass of magnesium (Mg) = 24.31 g/mol
Moles of magnesium = mass / molar mass = 16.3 g / 24.31 g/mol ≈ 0.671 mol
Next, calculate the moles of oxygen using the given mass:
Molar mass of oxygen (O2) = 32.00 g/mol
Moles of oxygen = mass / molar mass = 4.52 g / 32.00 g/mol ≈ 0.141 mol
According to the balanced chemical equation, the ratio of moles between magnesium and oxygen is 2:1. Therefore, we need 2 moles of magnesium for every 1 mole of oxygen.
The ratio of moles of magnesium to oxygen is 0.671 mol / 0.141 mol ≈ 4.76.
This means that we have an excess of magnesium since we only need a 2:1 ratio. In this case, oxygen will be the limiting reactant.
To determine the moles of magnesium oxide formed, we use the moles of the limiting reactant (oxygen) and the stoichiometry from the balanced equation:
Moles of magnesium oxide = 0.141 mol * (2 mol MgO / 1 mol O2) = 0.282 mol
Finally, calculate the grams of magnesium oxide formed using the molar mass:
Molar mass of magnesium oxide (MgO) = 40.31 g/mol
Grams of magnesium oxide = moles * molar mass = 0.282 mol * 40.31 g/mol ≈ 11.38 g
Therefore, approximately 11.38 grams of magnesium oxide will be formed.
In this reaction, oxygen is the limiting reactant and magnesium is the excess reactant.
To determine the grams of magnesium oxide formed in the reaction, we need to find the limiting reactant first. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed. The excess reactant is the one left over after the limiting reactant is completely used up.
To find the limiting reactant, we need to compare the amount of product that can be formed by each reactant. We can do this by using the balanced chemical equation for the reaction:
2Mg + O₂ → 2MgO
Looking at the equation, we can see that 2 moles of magnesium (Mg) reacts with 1 mole of oxygen (O₂) to produce 2 moles of magnesium oxide (MgO).
1. Calculate the number of moles for each reactant:
Moles of magnesium (Mg) = mass / molar mass
Moles of oxygen (O₂) = mass / molar mass
The molar mass of magnesium (Mg) is 24.31 g/mol, and the molar mass of oxygen (O₂) is 32 g/mol.
- Moles of magnesium (Mg) = 16.3 g / 24.31 g/mol = 0.671 moles
- Moles of oxygen (O₂) = 4.52 g / 32 g/mol = 0.141 moles
2. Divide the number of moles of each reactant by their respective stoichiometric coefficients from the balanced equation to find the mole ratio with respect to the product.
- Moles of magnesium oxide (MgO) from magnesium (Mg) = 0.671 moles / 2 = 0.336 moles
- Moles of magnesium oxide (MgO) from oxygen (O₂) = 0.141 moles / 1 = 0.141 moles
3. The reactant that produces the lesser amount of moles of magnesium oxide (MgO) is the limiting reactant. In this case, oxygen (O₂) produces fewer moles of magnesium oxide (MgO) compared to magnesium (Mg).
Therefore, oxygen (O₂) is the limiting reactant, and magnesium (Mg) is the excess reactant.
To calculate the grams of magnesium oxide (MgO) formed, we can use the limiting reactant (oxygen) and convert moles to grams using the molar mass of magnesium oxide (MgO), which is 40.31 g/mol.
- Grams of magnesium oxide (MgO) = moles of magnesium oxide (MgO) * molar mass of magnesium oxide (MgO)
- Grams of magnesium oxide (MgO) = 0.141 moles * 40.31 g/mol = 5.682 g
Therefore, 5.682 grams of magnesium oxide (MgO) will be formed when 16.3 grams of magnesium (Mg) reacts with 4.52 grams of oxygen (O₂).
Do you know how to solve simple stoichiometry problems; i.e., say 16.4 g Mg and all of the oxygen you need (instead of an amount listed). If you do the limiting reagent problem is a quick switch. Here is an example of a simple stoichiometry problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html