What is the actual amount of magnesium oxide produced when excess carbon
dioxide reacts with 42.8 g of magnesium metal? The percent yield of MgO(s)for
this reaction is 81.7%.
2Mg(s) . CO2(g) y 2MgO(s) . C(s)
To calculate the actual amount of magnesium oxide produced, we first need to calculate the theoretical yield of magnesium oxide using the given information and balanced equation.
Step 1: Convert the mass of magnesium metal to moles.
Given: Mass of magnesium metal = 42.8 g
Molar mass of magnesium (Mg) = 24.31 g/mol
Number of moles of magnesium = Mass of magnesium metal / Molar mass of magnesium
= 42.8 g / 24.31 g/mol
≈ 1.76 mol
Step 2: Use the balanced equation to determine the moles of magnesium oxide produced.
From the balanced equation, we see that 2 moles of magnesium (Mg) react to form 2 moles of magnesium oxide (MgO).
Number of moles of magnesium oxide produced = 1.76 mol
Step 3: Calculate the molar mass of magnesium oxide (MgO).
Molar mass of magnesium oxide (MgO) = 40.31 g/mol
Step 4: Calculate the theoretical yield of magnesium oxide.
Theoretical yield of magnesium oxide = Number of moles of magnesium oxide produced × Molar mass of magnesium oxide
= 1.76 mol × 40.31 g/mol
≈ 70.98 g
So, the theoretical yield of magnesium oxide is approximately 70.98 g.
Step 5: Calculate the actual amount of magnesium oxide produced using the percent yield.
Given: Percent yield = 81.7%
Actual yield of magnesium oxide = Percent yield × Theoretical yield
= 81.7% × 70.98 g
≈ 57.99 g
So, the actual amount of magnesium oxide produced is approximately 57.99 g.
To find the actual amount of magnesium oxide (MgO) produced, we first need to calculate the theoretical yield of MgO using stoichiometry and the given amount of magnesium metal (Mg).
From the balanced chemical equation:
2Mg(s) + CO2(g) → 2MgO(s) + C(s)
We can see that the ratio between Mg and MgO is 2:2, which means one mole of magnesium metal will react to give one mole of magnesium oxide. Therefore, the moles of magnesium (Mg) can be calculated using the molar mass of magnesium (24.31 g/mol).
Moles of Mg = mass (g) / molar mass (g/mol) = 42.8 g / 24.31 g/mol = 1.76 mol
Since the molar ratio between Mg and MgO is 1:1, the moles of MgO produced will also be 1.76 mol.
Now, to find the actual amount of MgO produced, we need to calculate its mass using the molar mass of magnesium oxide (40.31 g/mol).
Mass of MgO = moles of MgO × molar mass of MgO = 1.76 mol × 40.31 g/mol = 70.95 g
Therefore, the actual amount of magnesium oxide produced when excess carbon dioxide reacts with 42.8 g of magnesium metal is 70.95 grams.
Note: The given percent yield of 81.7% suggests that the actual yield is 81.7% of the theoretical yield. To find the theoretical yield, we divide the actual yield by the percent yield and multiply by 100.
Theoretical yield of MgO = actual yield (g) / percent yield = 70.95 g / 0.817 = 86.82 g