0.560g of potassium hydroxide is added to 25.0 cm^3 of 1.00 mol dm^-3 of hydrochloric acid.Find the limiting reagent & please show the working.THANKS IN ADVANCE.
1. Write the equation and balance it.
2. Set up an ICE chart.
moles KOH = grams/molar mass = 0.01
moles HCl = M x L = 0.025
.........KOH + HCl ==> H2O + KCl
initial..0.01...0.025....0.....0
Just by looking at this equation and the numbers you should be able to tell that KOH is the limiting reagent. It will form 0.01 mole H2O and 0.01 moles KCl and 0.025 - 0.01 = 0.015 moles HCl remain unreacted.
This is a relatively easy one because the coefficient for all of the reactants and products is one; if the ratio was not 1:1 for acid and base, the 1:1 comparison would not work.
To find the limiting reagent in a chemical reaction, you need to compare the stoichiometric quantities of the reactants and determine which one will be completely consumed, thereby limiting the amount of product that can be formed.
Let's start by writing the balanced chemical equation for the reaction between potassium hydroxide (KOH) and hydrochloric acid (HCl):
KOH + HCl → KCl + H2O
From the equation, we can see that the stoichiometric ratio between KOH and HCl is 1:1. This means that 1 mole of KOH reacts with 1 mole of HCl.
Now, let's calculate the amount of hydrochloric acid in moles:
Volume of HCl = 25.0 cm^3 = 25.0 × 10^-3 dm^3 (converted to dm^3)
Concentration of HCl = 1.00 mol dm^-3
Number of moles of HCl = Concentration × Volume
= 1.00 mol dm^-3 × 25.0 × 10^-3 dm^3
= 0.025 mol
Next, calculate the amount of potassium hydroxide in moles:
Mass of KOH = 0.560 g (given)
Now, we need to convert the mass of KOH to moles using its molar mass.
Molar mass of KOH = (atomic mass of K) + (atomic mass of O) + (atomic mass of H)
= (39.10 g/mol) + (16.00 g/mol) + (1.01 g/mol)
= 56.11 g/mol
Number of moles of KOH = Mass / Molar mass
= 0.560 g / 56.11 g/mol
= 0.00997 mol (approximately 0.010 mol)
After comparing the quantities of moles of KOH and HCl, we can see that the amount of KOH (0.010 mol) is smaller than the amount of HCl (0.025 mol). Therefore, KOH is the limiting reagent in this reaction.
I hope this helps! Let me know if you have any further questions.