Asked by ALIYAH
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.
Answers
Answered by
DrBob222
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.
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.
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