0.5 grms of the magnesium burnt in a closed vessel which contents 0.25grms oxygen.
1)which reaction limiting reagent?
2)find the mass of access reaction?
1. Write and balance the equation.
2Mg + O2 ==> 2MgO
2. Convert what you have to mols.
a. mols Mg = grams/atomic mass = ?
b. mols O2 = grams/molar mass = ?
3. Using the coefficients in the balanced equation, convert mols Mg and mols O2 to mols of MgO
a. You likely will find that the mols MgO proeduced is NOT the same for both Mg and O2.
b. The SMALLER of the two is the correct value to use. The reagent (Mg or O2) producing the smaller number is the limiting reagent (LR).
Now for the mass of the EXCESS (ER),
1. Using the coefficients from the LR, convert that to mols of the ER. That will be mols ER USED.
2. Convert mols ER used to grams. grams = mols x molar mass = ?
3. Subtract grams ER used from grams ER initially from the problem. That will be grams ER remaining. That is the excess.
Post your work if you get stuck.
To determine the limiting reagent and the mass of excess reactant, we need to compare the moles of magnesium and oxygen present and use stoichiometry to calculate the quantities involved.
1) To find the limiting reagent:
Step 1: Convert the masses of magnesium and oxygen to moles.
Moles of magnesium = mass of magnesium / molar mass of magnesium
Moles of magnesium = 0.5 g / 24.31 g/mol (molar mass of magnesium)
Moles of oxygen = mass of oxygen / molar mass of oxygen
Moles of oxygen = 0.25 g / 16.00 g/mol (molar mass of oxygen)
Step 2: Determine the mole ratio between magnesium and oxygen based on the balanced chemical equation.
The balanced equation for the reaction between magnesium and oxygen is:
2 Mg + O2 -> 2 MgO
From the equation, we can see that 2 moles of magnesium react with 1 mole of oxygen to produce 2 moles of magnesium oxide.
Step 3: Compare the moles of magnesium and oxygen to determine the limiting reagent.
Compare the moles of magnesium and oxygen using their respective mole ratios from the balanced equation. The reactant that has fewer moles is the limiting reagent.
In this case, the moles of magnesium is (0.5 g / 24.31 g/mol) = 0.0206 mol
The moles of oxygen is (0.25 g / 16.00 g/mol) = 0.0156 mol
Since there are fewer moles of oxygen, it is the limiting reagent.
2) To find the mass of excess reactant:
Step 1: Calculate the moles of the excess reactant.
Moles of excess reactant = moles of limiting reagent (oxygen) x (mole ratio of excess reactant from the balanced equation)
From the balanced equation, the mole ratio between oxygen and magnesium is 1:2.
Moles of excess reactant (magnesium) = (0.0156 mol oxygen) x (2 mol magnesium / 1 mol oxygen) = 0.0312 mol
Step 2: Convert the moles of excess reactant to mass.
Mass of excess reactant = moles of excess reactant x molar mass of excess reactant
Mass of excess reactant = 0.0312 mol x 24.31 g/mol (molar mass of magnesium)
Therefore, the mass of the excess reactant is 0.757 g (rounded to three decimal places).
In summary:
1) The limiting reagent is oxygen.
2) The mass of the excess reactant (magnesium) is 0.757 grams.