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1 year ago

Write ionic equations for the following: HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)

Chemistry

1 answer:

vodomira [7]1 year ago

4 0

Answer:

HCl(aq) + KOH(aq) ===> H2O(l) + KCl(aq)

Note the stoichiometry of the balanced equations shows us that HCl and KOH react in a 1:1 mole ratio. So, let us find moles of HCl and moles of KOH that are present:

moles HCl = 250.0 ml x 1 L/1000 ml x 0.25 mol/L = 0.06250 moles HCl

moles KOH = 200.0 ml x 1 L/1000 ml x 0.40 mol/L = 0.0800 moles KOH

You can see that there are more moles of KOH than there are of HCl, meaning that KOH is in excess and after neutralizing all of the HCl, the solution will be left with excess KOH making the pH > 7 = BASIC

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The volume of H₂ (g) obtained is 22.4L

Explanation:

First of all, think the reaction:

2HCl (aq) + Zn (s) → ZnCl₂ (aq) + H₂ (g)

You have to add a 2, in the HCl to get ballanced.

Now we should know how many moles of each reactant, do we have.

Volume . Molarity = moles

Notice that volume is in mL, so I must convert to L.

275 mL = 0.275L

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50 g / 65.41 g/m = 0.764 moles

Ratio between reactants is 2:1, so I need the double of moles of HCl to react, and a half moles of Zn to react.

My limiting reactant is the HCl, for 0.764 moles of Zinc, I need 1.528 (0.764 .2) of HCl, and I only have 0.2 moles.

Ratio between HCl and H₂ is 1:1, so 0.2 HCl make 0.2 moles of gas

Now apply the Ideal Gas Law, to find out the volume

P. V = n . R . T

2 atm . V = 0.2 mol . 0.08206L atm/K mol . 273K

V = (0.2 mol . 0.08206L atm/K mol . 273K ) / 2 atm

V = 2.24 L

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