If 12.01 g of Mg metal and 6.56 g of O2 gas are reacted,
1) How many moles of MgO can be produced from the given mass of Mg?
2) How many moles of MgO can be produced from the given mass of O2?
3) From the above calculations, determine which of the reactants is the limiting reagent.
Recall the balanced reaction is:
2 Mg (s) + O2 (g) 2 MgO (s)
Enter the number of moles of MgO from 12.01 of Mg here:
Enter the number of moles of MgO from 6.56 of O2 here:
The limiting reagent is?: Mg O2
2Mg + O2 ==> 2MgO
mols Mg = grams/atomic mass = 12.01/24.3 = ?
mols O2 = grams/molar mass = 6.56/32 = ?
Using the coefficients in the balanced equation, convert mols Mg to mols MgO produced. That's #1.
#2. Convert mols O2 to mols MgO.
#3. In limiting reagent problems the SMALLER value will be the correct number of mols. The reagent responsible for the value is called the limiting reagent. It will be used up completely and the other one is the excess reagent.
To answer these questions, we need to use the concept of stoichiometry and the given balanced chemical equation. Let's go through each question step by step:
1) How many moles of MgO can be produced from the given mass of Mg?
To find the number of moles of MgO that can be produced, we first need to calculate the number of moles of Mg. We can do this by dividing the given mass of Mg (12.01 g) by the molar mass of Mg, which is 24.31 g/mol:
Number of moles of Mg = Mass of Mg / Molar mass of Mg
= 12.01 g / 24.31 g/mol
= 0.4944 mol
Since the balanced chemical equation tells us that 2 moles of Mg reacts to produce 2 moles of MgO, we can conclude that:
Number of moles of MgO = Number of moles of Mg
= 0.4944 mol
So, 0.4944 moles of MgO can be produced from the given mass of Mg.
2) How many moles of MgO can be produced from the given mass of O2?
Similar to the previous calculation, we need to find the number of moles of O2. We can do this by dividing the given mass of O2 (6.56 g) by the molar mass of O2, which is 32.00 g/mol:
Number of moles of O2 = Mass of O2 / Molar mass of O2
= 6.56 g / 32.00 g/mol
= 0.205 mol
Since the balanced chemical equation tells us that 1 mole of O2 reacts to produce 2 moles of MgO, we can conclude that:
Number of moles of MgO = 2 * Number of moles of O2
= 2 * 0.205 mol
= 0.410 mol
So, 0.410 moles of MgO can be produced from the given mass of O2.
3) From the above calculations, determine which of the reactants is the limiting reagent.
To determine the limiting reagent, we compare the stoichiometric ratios of Mg to MgO and O2 to MgO. Whichever ratio is smaller corresponds to the limiting reagent.
The stoichiometric ratio of Mg to MgO is 2:2, which means 2 moles of Mg are needed to produce 2 moles of MgO.
The stoichiometric ratio of O2 to MgO is 1:2, which means 1 mole of O2 is needed to produce 2 moles of MgO.
From our calculations, we found that we have 0.4944 moles of Mg and 0.410 moles of O2. Since we have less O2 compared to Mg, the O2 is the limiting reagent.
So, the limiting reagent is O2.
Please enter:
The number of moles of MgO from 12.01 g of Mg = 0.4944
The number of moles of MgO from 6.56 g of O2 = 0.410
The limiting reagent is O2.