This problem of the limiting reactant is:
Given the reactant amounts specified in each chemical reaction, dtermine the limiting reactant in each case:
(I'm just going to take one problem out of the bunch}
HCl + NaOH yeilds NaCl + H2O
2.00 2.50
mol mol
(answers have to be in 3 significant figures)
2.50molNaOH x 1.00molHCl = 2.50 molof
1.00molNaOH HCl
So now we have to find the amount of moles of excess reactant that remains and that's where I get stuck
There is a short way and a long way. I like the long way because its easier to explain. After you are acquainted with the problems of this type you can switch to the short way.
1. You have the equation.
Step 2. Convert grams to mols but you have mol now.
Step 3a. Use the equation to convert mols of HCl to mols of either water or NaCl. Let's choose NaCl.
2.00 mols HCl x (1 mol NaCl/1 mol HCl) = 2.00 mols NaCl produced.
Step 3b. Do the same thing for NaOH.
2.50 mols NaOH x (1 mol NaCl/1 mol NaOH) = 2.50 mols NaCl produced.
Step 3c. Obviously, both answers can't be correct. The amount of product produced is always the smaller of the two numbers. In this case, we will produce 2.00 mols NaCl AND THIS MAKES HCl the limiting reagent AND IT MAKES NaCl the excess reagent.
Step 3d. Convert mols limiting reagent to mols of the excess reagent.
2.00 mols HCl x (1 mol NaOH/1 mol HCl) = 2.00 mols NaOH used.
Step 3e. How much NaOH remains?
2.50 mol initially - 2.00 mol used = 0.50 mol NaOH remains unreacted. And you are not allowed three significant figures in this case because neither HCl nor NaOH are given past the hundreths place.
The short way.
2.00 mol HCl reacts with 2.00 mols NaOH leaving 2.50 - 2.00 mol = 0.50 mol NaOH unreacted.
BTW, notice the correct spelling of yields. Or place an --> or ===> there.
Thanks sorry about the mistake in spelling
No problem! Spelling mistakes happen. Let me summarize the steps to determine the limiting reactant in this problem:
1. Start with the balanced chemical equation: HCl + NaOH yields NaCl + H2O.
2. Convert the given amounts of reactants (in this case, 2.00 mol HCl and 2.50 mol NaOH) to the corresponding amount of product using stoichiometry. For example, for HCl: 2.00 mol HCl x (1 mol NaCl/1 mol HCl) = 2.00 mol NaCl produced.
3. Repeat the same calculation for the other reactant (NaOH): 2.50 mol NaOH x (1 mol NaCl/1 mol NaOH) = 2.50 mol NaCl produced.
4. Compare the calculated amounts of product. The smaller value represents the maximum amount of product that can be formed. In this case, 2.00 mol NaCl is the maximum amount produced, so HCl is the limiting reactant.
5. To find the amount of excess reactant remaining, convert the moles of the limiting reactant to the moles of the excess reactant. For example, for HCl: 2.00 mol HCl x (1 mol NaOH/1 mol HCl) = 2.00 mol NaOH used.
6. Finally, subtract the moles of NaOH used from the initial moles of NaOH to find the amount of NaOH remaining: 2.50 mol NaOH initially - 2.00 mol NaOH used = 0.50 mol NaOH remains unreacted.
The short way to determine the limiting reactant is to directly compare the mole ratios of the reactants in the balanced equation. In this case, since both reactants have a 1:1 mole ratio with NaCl, the limiting reactant can be determined by comparing the initial moles of each reactant. The reactant with the smaller initial moles is the limiting reactant, and the difference between the initial moles of the limiting reactant and the initial moles of the other reactant gives you the moles of excess reactant remaining.