What mass of iodine would react with 48 grams of magnesium to make magnesium iodide? mg + i2 -> mgi2

1) Chemical equation:

Mg + I2 ---> Mg I2

2) molar ratios

1 mol Mg : 1 mol I2 : 1 mol Mg I2

3) Calculate the number of moles of Mg in 48 grams

Atomic mass of Mg: 24.3 g/mol

number of moles = mass in grams / atomic mass = 48 g / 24.3 g/mol = 1.975 mol Mg

4) Use a proportion with the molar ratios

1 mol I2 / 1 mol Mg = x / 1.975 mol Mg =>

=> x = 1 mol I2 * 1.975 mol Mg / 1 mol Mg = 1.975 mol I2

5) Convert 1.975 mol I2 to grams

molar mass of I2 = 2 * 126.9 g/mol = 253.8 g/mol

mass = number of moles * molar mass = 1.975 mol * 253.8 g/mol = 501.255 g

501 grams

What’s the question

Hello! data: p (pressure) = 1.25 atmv (volume) = ? (in l)t (temperature) = 298 kn (number of mols) = 0.900 molesr (gas constant) = 0.082 (atm*l/mol*k)apply the data to the clapeyron equation (ideal gas equation), see: i hope this . =)