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

You might be interested in

Someone help me plz :))) brainliest

SOVA2 [1]

Answer:

Option 2

Explanation:

3 0

3 years ago

Wat two common uses for Gallium??

Ann [662]

Answer:

mirrors and metal thermometers

Explanation:

ooga booga

6 0

3 years ago

Read 2 more answers

What are 5 renewable energy Resources and why are they better for the society and the environment

Ket [755]

Answer:

hydroelectric , hydrogen fuel cells , solar power , geothermal , wind power

Explanation:

all of these has little to no waste being produced and have little to no impact on the environment .

5 0

3 years ago

Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K

Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

$MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-$

And the undergoing chemical reaction:

$MgCl_2+2NaF\rightarrow MgF_2+2NaCl$

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

$n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2$