How many grams of magnesium bromide are produced if 16 g of bromine gas and 4 g of magnesium are used?

Question

And...

Which of the reactant is limited?

Which of the reactant is excessive?

Answer 1

First you'll want to figure out the reaction.

The unbalanced equation is as follows:

_Br₂ + _Mg → _MgBr₂

Balancing the equation yields a 1:1:1 ratio so the balanced equation is:

Br₂ + Mg → MgBr₂

Now let's figure out the amount of each reactant in moles

The molar mass of Br₂ is 2 * 79.9 g / mol = 159.8 g / mol

So the moles of bromine gas equals:

16 g Br₂ / (159.8 g / mol) = 0.10 mol Br₂

The molar mass of Magnesium is 24.3 g / mol

So the moles of magnesium equals:

4 g Mg / (24.3 g / mol) = 0.16 mol Mg

Since the molar ratio of Mg : Br₂ is 1:1, and there are more moles of Mg than Br₂, Br₂ is the limiting reagent, and Mg is in excess.

Now, to find the amount of MgBr₂ produced, simply take the moles of Br₂ (the limiting reagent) and convert to moles of MgBr₂ as follows:

The molar mass of MgBr₂ is 24.3 g / mol (for Mg) + 2 * 79.9 g / mol (for Br) = 184.1 g / mol

0.10 mol Br₂ * (1 mol MgBr₂ / 1 mol Br₂) * (184.1 g MgBr₂ / 1 mol MgBr₂) = 18.41 g of MgBr₂

You will need to check the significant figures here, technically you gave one significant figure for the mass of magnesium (4 g), so the answer will only have 1 significant figure, which in this case would yield 20 g.