2Mg(s) + O2(g) 2MgO(s)
If 9.92 g of Mg reacts with 8.52 g of O2. How many grams of MgO could be produced?
I am just very confused and need an answer so that I can look back on how you got there . . .
This is a limiting reagent (LR) problem because amounts are given for BOTH reactants. You need to find the arrow key and use it; otherwise we don't know where the reactants end and the products start.
2Mg(s) + O2(g) 2MgO(s)
mols Mg = grams/molar mass
mols O2 = grams/molar mass
Using the coefficients in the balanced eqwuation, convert mols Mg to mols MgO.
Do the same and convert mols O2 to mols MgO.
It is likely that the two values for mols MgO will not agree which means one of them is wrong; the correct value in LR problems is ALWAYS the smaller one and the reagent producing that number is the LR.
Use the smaller number and convert to grams. g MgO = mols MgO x molar mass MgO.
To determine the amount of MgO that could be produced in the given reaction, we need to use the concept of stoichiometry.
1. Start by calculating the number of moles for each reactant.
The molar mass of Mg (magnesium) is 24.31 g/mol, so we can calculate the number of moles of Mg:
moles of Mg = mass of Mg / molar mass of Mg
moles of Mg = 9.92 g / 24.31 g/mol
moles of Mg ≈ 0.41 mol
The molar mass of O2 (oxygen) is 32.00 g/mol, so we can calculate the number of moles of O2:
moles of O2 = mass of O2 / molar mass of O2
moles of O2 = 8.52 g / 32.00 g/mol
moles of O2 ≈ 0.27 mol
2. Determine the limiting reactant.
To find the limiting reactant, we need to compare the mole ratio between Mg and O2 in the balanced equation.
From the balanced equation, we can see that the mole ratio between Mg and O2 is 2:1. This means that for every 2 moles of Mg, we need 1 mole of O2 to react completely and form MgO.
Since the ratio of actual moles is less than the ratio in the balanced equation, O2 is the limiting reactant because there is less of it.
3. Calculate the moles of MgO produced.
Since O2 is the limiting reactant, we can use its moles to determine the moles of MgO formed. According to the balanced equation, 2 moles of Mg react to produce 2 moles of MgO.
moles of MgO = moles of O2 (limiting reactant) × (moles of MgO / moles of O2) from the balanced equation
moles of MgO = 0.27 mol × (2 mol MgO / 1 mol O2)
moles of MgO = 0.54 mol
4. Convert moles of MgO to grams.
Now, we can calculate the mass of MgO produced using its molar mass.
mass of MgO = moles of MgO × molar mass of MgO
mass of MgO = 0.54 mol × 40.31 g/mol (molar mass of MgO)
mass of MgO ≈ 21.79 g
Therefore, approximately 21.79 grams of MgO could be produced from the given reactants.