Consider this real reaction, and answer the questions based on it: TiO2 (s) + HCl (aq) ==> H2O (l) + TiCl4 (aq)

Question

1) Based on this balanced equation, if the amounts used are 3 mol TiO2 and 5 mol HCl, Which one is the limiting reactant?

2) How many moles of each reactant are leftover?

3) How many moles of each product are produced?

I understand the balanced equation is TiO2s (s) + 4HCL (aq) ---> 2H2O (l) + TiCl4 (aq)

But I do not know what I need to do next to answer these questions. I started adding up their molar masses and became confused as to what to do or if that was the right step. Thank you in advance for your time I really appreciate it!

Answer 1

I will omit the phases.
TiO2 + 4HCl ---> 2H2O + TiCl4

Since this is a limiting reagent problem (LR), I always do these the long way; i.e., calculate mols product formed from BOTH and take the smaller value.

First 3 mol TiO2.
3 mol TiO2 x (1 mol TiCl4/1 mol TiO2) = 3 mol TiCl4 produced if we had all of the HCl needed.

Next 5 mols HCl.
5 mol HCl x (1 mol TiCl4/4 mols HCl) = 5 x 1/4 = 1.25 mols TiCl4 produced if we had all of the TiO2 needed.

In LR problems the smaller number ALWAYS wins; therefore, 1.25 mols TiO2 is formed and HCl is the LR.

2. How much is left. All of these are just conversions of 1 reactant into another. For the HCl, we had 5, we used all of it, we have zero left.
TiO2. How much TiO2 did we use? That's 5 mols HCl x (1 mol TiO2/4 mols HCl) = 5 x 1/4 = 1.25 mols TiO2 used. We had 3.0 initiall; therefore, we have 3.0-1.25 left.

3. Use the LR of HCl and solve for mols TiO2 and mols H2O producd. I've done the TiO2 above as a part of finding the LR.

Post your work if you get stuck.