Question

Cobalt (III) bromide reacts with lithium hydroxide to produce lithium bromide and cobalt (III) hydroxide. Please answer the questions below for this reaction: 1. How many moles of cobalt (III) bromide were reacted if 6.5 moles of lithium bromide were produced? 2. If 17.9 moles of cobalt (III) hydroxide were produced how many moles of lithium hydroxide were reacted?

Answer 1

Let's balance the chemical equation for the reaction first to understand the stoichiometry:

Cobalt (III) bromide + Lithium hydroxide → Lithium bromide + Cobalt (III) hydroxide

The balanced equation is: 2 CoBr3 + 6 LiOH → 6 LiBr + Co(OH)3

Now, we can use the coefficients from the balanced equation.

How many moles of cobalt (III) bromide were reacted if 6.5 moles of lithium bromide were produced?

From the balanced equation, the ratio of CoBr3 to LiBr is 2:6, which simplifies to 1:3.

So, if 6.5 moles of LiBr were produced, the moles of CoBr3 reacted would be:

Moles of CoBr3 = (6.5 moles LiBr) * (1 mole CoBr3 / 3 moles LiBr) = 2.17 moles CoBr3

If 17.9 moles of cobalt (III) hydroxide were produced, how many moles of lithium hydroxide were reacted?

From the balanced equation, the ratio of LiOH to Co(OH)3 is 6:1.

So, if 17.9 moles of Co(OH)3 were produced, the moles of LiOH reacted would be:

Moles of LiOH = (17.9 moles Co(OH)3) * (6 moles LiOH / 1 mole Co(OH)3) = 107.4 moles LiOH