Consider this real reaction, and answer the questions based on it: TiO2 (s) + HCl (aq) ==> H2O (l) + TiCl4 (aq)
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!
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.
To determine the limiting reactant and the amounts of reactants and products, you need to use stoichiometry. Here's how you can approach each question:
1) Determine the moles of each reactant used:
- TiO2: 3 mol
- HCl: 5 mol
Since the stoichiometric ratio between TiO2 and HCl in the balanced equation is 1:4, you can calculate the moles of HCl required to completely react with 3 mol of TiO2 using this ratio:
3 mol TiO2 x (4 mol HCl / 1 mol TiO2) = 12 mol HCl
Since you have only 5 mol of HCl, which is less than the required 12 mol, HCl is the limiting reactant.
2) Calculate the moles of each reactant leftover:
To find the moles of leftover TiO2 or HCl, subtract the moles used from the initial amount:
- Moles of TiO2 leftover = 3 mol - 3 mol = 0 mol
- Moles of HCl leftover = 5 mol - 12 mol = -7 mol
Note that the negative value for HCl indicates that it is completely consumed, and there are none left.
3) Determine the moles of each product produced:
Based on the stoichiometry of the balanced equation, 1 mol of TiO2 reacts with 4 mol of HCl to produce 2 mol of H2O and 1 mol of TiCl4.
Since HCl is the limiting reactant, we use its moles to calculate the moles of products:
- Moles of H2O produced = 5 mol HCl x (2 mol H2O / 4 mol HCl) = 2.5 mol H2O
- Moles of TiCl4 produced = 5 mol HCl x (1 mol TiCl4 / 4 mol HCl) = 1.25 mol TiCl4
Therefore, 2.5 mol of H2O and 1.25 mol of TiCl4 are produced.
Remember to always use the balanced equation, mole ratios, and stoichiometry to solve these types of problems.
To determine the limiting reactant and the amount of reactants left over, you need to compare the ratio of the moles of the reactants to the stoichiometric ratio in the balanced equation.
1) To find the limiting reactant, compare the moles of TiO2 to the moles of HCl, using the stoichiometric ratio from the balanced equation: 1 mol TiO2 reacts with 4 mol HCl.
Given:
- Moles of TiO2 = 3 mol
- Moles of HCl = 5 mol
To determine the limiting reactant, you need to find the mole ratio between TiO2 and HCl. Divide the moles of each reactant by their respective stoichiometric coefficient (the number in front of the reactant in the balanced equation):
- Moles of TiO2 / Stoichiometric coefficient of TiO2 = 3 mol / 1 = 3 mol
- Moles of HCl / Stoichiometric coefficient of HCl = 5 mol / 4 = 1.25 mol
Comparing these ratios, you can see that TiO2 is present in excess with 3 mol, while HCl is the limiting reactant with only 1.25 mol.
2) To calculate the moles of the reactants left over, subtract the moles of the limiting reactant from the initial moles of each reactant:
- Moles of TiO2 left over = Initial moles of TiO2 - Moles of TiO2 consumed
= 3 mol - 3 mol (since TiO2 is in excess)
= 0 mol
- Moles of HCl left over = Initial moles of HCl - Moles of HCl consumed
= 5 mol - 1.25 mol
= 3.75 mol
Therefore, there are 0 mol of TiO2 left over, and 3.75 mol of HCl left over.
3) To determine the moles of each product produced, you use the stoichiometric coefficients from the balanced equation:
- Moles of H2O produced = Stoichiometric coefficient of H2O * Moles of limiting reactant
= 2 * 1.25 mol (since HCl is the limiting reactant)
= 2.5 mol
- Moles of TiCl4 produced = Stoichiometric coefficient of TiCl4 * Moles of limiting reactant
= 1 * 1.25 mol (since HCl is the limiting reactant)
= 1.25 mol
Therefore, 2.5 mol of H2O and 1.25 mol of TiCl4 are produced.
Remember, these calculations are based on the stoichiometry of the balanced equation and the given amounts of reactants.