Question

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

Answer 1

48 grams of magnesium will react to 501.255 gm of iodine to make magnesium iodide.

How is it carried out?

Following reaction would carry out

$Mg + I_{2} -- > MgI_{2}$

So, the molar ratio of the equation would be

1 mol Mg  : 1 mol $I_{2}$ : 1 mol $MgI_{2}$

Now finding how many grams of iodine needed to reaction with 48 gm of Magnesium (Mg).

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

Proportion Ratio

1 mol $I_{2}$ gives 1 mol Mg

x mol $I_{2}$ gives 1.975 mol Mg

Therefore,

1 mol $I_{2}$ / 1 mol Mg = x / 1.975 mol Mg

x = 1 mol $I_{2}$ * 1.975 mol Mg / 1 mol Mg

x=1.975 mol $I_{2}$

Now converting 1.975 mol $I_{2}$ into grams

molar mass of  $I_{2}$ = 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

Therefore 501.255 g iodine would react with 48 grams of magnesium to make magnesium iodide

To learn more about this kind of problems:

brainly.com/question/13013091

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