Calculate the mass of Mg metal requied to consume 2560g of CO2 in the reaction 2Mg+CO2 → 2MgO + C
2Mg + CO2 → 2MgO + C
mols CO2 = g CO2/molar mass CO2 = ?
Using the coefficients in the balanced equation convert mols CO2 to mols Mg needed. That's mols Mg = mols CO2 x (2 mol Mg/1 mol CO2)
Now convert mols Mg to grams Mg. That's g Mg = mols Mg x molar mass Mg
Post your work if you get stuck.
Numerical problem
Well, let's break it down, shall we? According to the balanced equation, 2 moles of Mg react with 1 mole of CO2. To calculate the mass of Mg required, we need to convert the given mass of CO2 to moles first.
The molar mass of CO2 is approximately 44 g/mol. So, to convert 2560 g of CO2 to moles, we divide the mass by the molar mass:
2560 g CO2 / 44 g/mol CO2 = 58.18 mol CO2
Now, we know that 2 moles of Mg reacts with 1 mole of CO2. Therefore, the moles of Mg required can be calculated by multiplying the moles of CO2 by the ratio:
58.18 mol CO2 * (2 mol Mg / 1 mol CO2) = 116.36 mol Mg
Finally, we can convert moles of Mg to grams by multiplying by the molar mass of Mg, which is approximately 24 g/mol:
116.36 mol Mg * 24 g/mol Mg = 2792.64 g Mg
Therefore, approximately 2792.64 grams of Mg metal would be required to consume 2560 grams of CO2 in the given reaction. That's a lot of Mg!
To find the mass of Mg required to consume 2560g of CO2, we can use the stoichiometry of the reaction.
The balanced equation is:
2Mg + CO2 → 2MgO + C
From this equation, we can see that 1 mole of Mg reacts with 1 mole of CO2. The molar mass of Mg is 24.31 g/mol.
Step 1: Calculate the moles of CO2.
To do this, divide the given mass of CO2 by its molar mass:
2560g CO2 / 44.01 g/mol = 58.171 mol CO2
Step 2: Use the stoichiometry of the reaction to find the moles of Mg.
From the balanced equation, we know that 2 moles of Mg react with 1 mole of CO2. Therefore, the number of moles of Mg is half of the moles of CO2:
58.171 mol CO2 × (2 mol Mg / 1 mol CO2) = 116.342 mol Mg
Step 3: Calculate the mass of Mg.
To do this, multiply the number of moles of Mg by its molar mass:
116.342 mol Mg × 24.31 g/mol = 2831.49 g
Therefore, the mass of Mg required to consume 2560g of CO2 in the given reaction is approximately 2831.49 grams.
To calculate the mass of Mg metal required to consume 2560g of CO2, we need to use the stoichiometry of the balanced chemical equation.
The balanced chemical equation is:
2Mg + CO2 → 2MgO + C
From the equation, we can see that for every 1 mole of CO2, we need 2 moles of Mg. This means the molar ratio between Mg and CO2 is 2:1.
To find the mass of Mg required, we follow these steps:
1. Determine the molar mass of CO2 and Mg.
2. Convert the mass of CO2 to moles using its molar mass.
3. Use the mole ratio from the balanced equation to find the moles of Mg.
4. Convert the moles of Mg to mass using its molar mass.
Let's calculate step by step:
Step 1: Determine the molar mass of CO2 and Mg.
The molar mass of CO2 (carbon dioxide):
C: 12.01 g/mol
O: 16.00 g/mol
Molar mass of CO2 = (12.01 g/mol) + 2(16.00 g/mol) = 44.01 g/mol
The molar mass of Mg:
Mg: 24.31 g/mol
Step 2: Convert the mass of CO2 to moles using its molar mass.
Moles of CO2 = mass of CO2 / molar mass of CO2
Moles of CO2 = 2560 g / 44.01 g/mol ≈ 58.171 mol
Step 3: Use the mole ratio from the balanced equation to find the moles of Mg.
From the balanced equation, 2 moles of Mg reacts with 1 mole of CO2.
Therefore, Moles of Mg = 2 × Moles of CO2
Moles of Mg = 2 × 58.171 mol = 116.342 mol
Step 4: Convert the moles of Mg to mass using its molar mass.
Mass of Mg = Moles of Mg × Molar mass of Mg
Mass of Mg = 116.342 mol × 24.31 g/mol ≈ 2,830.56 g
Therefore, approximately 2830.56 grams of Mg metal is required to consume 2560 grams of CO2.