Magnesium can be produced by reaction of magnesium oxide (MgO) with silicon (Si). The by-product of the reaction is magnesium orthosilicate (Mg2SiO4). A reactor is charged with 183.0 kg of MgO and 116.0 kg Si. Assume the reaction goes to completion. Express your answers in kg.
Enter the number of kilograms of MgO remaining:
Enter the number of kilograms of Si remaining:
Enter the number of kilograms of Mg present:
Enter the number of kilograms of Mg2SiO4 present:
Another limiting reagent problem.
To solve this problem, we need to understand the reaction and the stoichiometry involved.
The balanced chemical equation for the reaction is:
3MgO + Si → Mg2SiO4 + 2Mg
From the balanced equation, we can deduce that for every 3 moles of MgO, we obtain 2 moles of Mg2SiO4 and 2 moles of Mg.
1 mole of MgO has a molar mass of (24.31 g/mol + 16.00 g/mol) = 40.31 g/mol.
1 mole of Si has a molar mass of 28.09 g/mol.
Given:
Mass of MgO = 183.0 kg
Mass of Si = 116.0 kg
To find the number of moles of MgO and Si present, we can divide their masses by their respective molar masses:
Moles of MgO = Mass of MgO / Molar mass of MgO
Moles of Si = Mass of Si / Molar mass of Si
Moles of MgO = 183.0 kg / (40.31 g/mol)
Moles of Si = 116.0 kg / (28.09 g/mol)
Now, let's find the limiting reactant - the reactant that will be completely consumed and determine the theoretical yields.
Using the mole ratio from the balanced equation, we convert the moles of MgO and Si to moles of Mg2SiO4 and Mg, respectively.
Moles of Mg2SiO4 = Moles of MgO * (2 moles of Mg2SiO4 / 3 moles of MgO)
Moles of Mg = Moles of Si * (2 moles of Mg / 1 mole of Si)
To find the actual masses of Mg2SiO4 and Mg produced, we multiply their respective moles by their molar masses:
Mass of Mg2SiO4 = Moles of Mg2SiO4 * (Molar mass of Mg2SiO4)
Mass of Mg = Moles of Mg * (Molar mass of Mg)
Finally, to find the remaining mass of MgO and Si, we subtract the moles of MgO and Si consumed from the initial masses:
Remaining mass of MgO = Mass of MgO - (Moles of MgO consumed * Molar mass of MgO)
Remaining mass of Si = Mass of Si - (Moles of Si consumed * Molar mass of Si)
Now let's plug in the values and calculate:
Molar mass of MgO = 40.31 g/mol
Molar mass of Si = 28.09 g/mol
Molar mass of Mg2SiO4 = (24.31 * 2 + 28.09 + 16.00 * 4) g/mol = 140.71 g/mol
Molar mass of Mg = 24.31 g/mol
Moles of MgO = 183.0 kg / 40.31 g/mol ≈ 4534.06 mol
Moles of Si = 116.0 kg / 28.09 g/mol ≈ 4126.34 mol
Moles of Mg2SiO4 = 4534.06 mol * (2 mol Mg2SiO4 / 3 mol MgO) ≈ 3022.71 mol
Moles of Mg = 4126.34 mol * (2 mol Mg / 1 mol Si) ≈ 8252.68 mol
Mass of Mg2SiO4 = 3022.71 mol * 140.71 g/mol ≈ 425865.16 g ≈ 425.87 kg
Mass of Mg = 8252.68 mol * 24.31 g/mol ≈ 200858.03 g ≈ 200.86 kg
Remaining mass of MgO = 183.0 kg - (3022.71 mol * 40.31 g/mol) ≈ -4593.21 kg (negative value implies that MgO is entirely consumed)
Remaining mass of Si = 116.0 kg - (4126.34 mol * 28.09 g/mol) ≈ 0.0 kg
Now we can summarize the results:
The number of kilograms of MgO remaining: 0.0 kg
The number of kilograms of Si remaining: 0.0 kg
The number of kilograms of Mg present: 200.86 kg
The number of kilograms of Mg2SiO4 present: 425.87 kg