Calculate the mass of Mg metal required to consume 2560g of CO2 in the reaction. 2Mg+CO2_2MgO+C
No of moles of CO2=58.18moles
No of moles of Mg=116.rumpled
Mass of Mg=2828.12g
1. mols CO2 = grams/molar mass
2. Using the coefficients in the balanced equation, convert mols CO2 to mols Mg. That will be mols CO2 x (2 mol Mg/1 mol CO2) = mols CO2 x (2/1) = ?
3. Now convert mols Mg to grams. g = mols x atomic mass.
Chemistry
To calculate the mass of Mg metal required to consume 2560g of CO2, we need to use the stoichiometry of the reaction. The balanced chemical equation for the reaction is:
2Mg + CO2 -> 2MgO + C
From the equation, we can see that 2 moles of Mg reacts with 1 mole of CO2. We need to convert the given mass of CO2 to moles, and then use the stoichiometric ratio to determine the moles of Mg required.
First, let's find the molar mass of CO2:
Molar mass of CO2 = (12.01 g/mol for carbon) + (2 * 16.00 g/mol for oxygen) = 44.01 g/mol
Now, we can convert the given mass of CO2 to moles using the molar mass:
Moles of CO2 = mass of CO2 / molar mass of CO2
= 2560 g / 44.01 g/mol
≈ 58.156 mol
The stoichiometric ratio tells us that 2 moles of Mg react with 1 mole of CO2. Therefore, the number of moles of Mg required can be calculated as:
Moles of Mg = 2 * Moles of CO2
= 2 * 58.156 mol
= 116.312 mol
Finally, we can convert the moles of Mg to mass using the molar mass of Mg:
Mass of Mg = Moles of Mg * Molar mass of Mg
= 116.312 mol * (24.31 g/mol for Mg)
≈ 2828.94 g
Therefore, the mass of Mg metal required to consume 2560g of CO2 in the reaction is approximately 2828.94 grams.
Calculate the mass of Mg metal required to consume 2560g of CO2 in the reaction C 12,Mg24 ,O16
Well, let's have a laugh while we tackle this problem!
First, let's calculate the molar mass of CO2. Carbon (C) has an atomic mass of 12.01 g/mol, and oxygen (O) has an atomic mass of 16.00 g/mol. Since CO2 has one carbon atom and two oxygen atoms, we can add them up to find that the molar mass is 12.01 g/mol + (16.00 g/mol x 2) = 44.01 g/mol.
Now, the balanced equation tells us that 2 moles of Mg react with 1 mole of CO2 to produce 2 moles of MgO and 1 mole of CO2.
Since we know the molar mass of CO2 is 44.01 g/mol, we can calculate the number of moles of CO2 we have by dividing the mass of CO2 given (2560 g) by its molar mass:
2560 g CO2 / 44.01 g/mol CO2 = 58.17 mol CO2
According to the stoichiometry, 2 moles of Mg react with 1 mole of CO2. Therefore, we need double the number of moles of Mg compared to CO2. So, the number of moles of Mg needed is:
2 x 58.17 mol CO2 = 116.34 mol Mg
Finally, we can use the molar mass of Mg (24.31 g/mol) to find the mass of Mg needed:
116.34 mol Mg x 24.31 g/mol Mg ≈ 2826.67 g Mg
So, approximately 2826.67 grams of Mg metal are required to consume 2560 grams of CO2 in the given reaction.
Now that's a whole lot of magnesium! I hope you enjoyed this chemical comedy routine!