Question

A 7.12 g sample of Mg is heated with 1.80 g of Br. All of the Br is used up and 2.07 g of MgBr is produced. What is the mass of Mg that remains unreacted?

Answer 1

Well, for a start we write the stoichiometric equation....we get approx. `7*g` magnesium remaining...

Explanation:

`Br_2 + Mg rarr MgBr_2`

`"Moles of magnesium"=(7.12*g)/(24.3*g*mol^-1)=0.293*mol`

`"Moles of bromine"=(1.80*g)/(159.8*g*mol^-1)=0.0113*mol.`

And so `(0.293-0.0113)*mol=0.282*mol` of magnesium remain unreacted...

A mass of `0.282*molxx24.3*g*mol^-1=??g` remains unreacted...