How many grams of MgCl2 would be required to make 440 g of magnesium ? The other product is chlorine.
To determine the number of grams of MgCl2 required to make 440 grams of magnesium, we first need to find the molar mass of magnesium chloride (MgCl2).
The molar mass of MgCl2 can be calculated by summing the atomic masses of its constituent elements. The atomic mass of magnesium (Mg) is 24.31 g/mol, and the atomic mass of chlorine (Cl) is 35.45 g/mol. Since there are two chlorine atoms in each molecule of MgCl2, we multiply the atomic mass of chlorine by 2.
Molar mass of MgCl2 = (1 × atomic mass of Mg) + (2 × atomic mass of Cl)
= (1 × 24.31 g/mol) + (2 × 35.45 g/mol)
= 24.31 g/mol + 70.9 g/mol
= 95.21 g/mol
Now, we know that one mole of MgCl2 has a mass of 95.21 grams. To find the number of moles of MgCl2 required to make 440 grams of magnesium, we can use the following formula:
Number of moles = Mass / Molar mass
Number of moles of MgCl2 = 440 g / 95.21 g/mol
≈ 4.62 moles
Finally, to convert the number of moles of MgCl2 into grams, we can use the molar mass:
Mass of MgCl2 = Number of moles × Molar mass
= 4.62 moles × 95.21 g/mol
≈ 439.4 grams
Therefore, approximately 439.4 grams of MgCl2 would be required to make 440 grams of magnesium.