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3 years ago

What’s the total ionic equation for NaOH reacting with HCl

Chemistry

1 answer:

exis [7]3 years ago

6 0

Answer:

Na⁺ (aq) + OH⁺ (aq) + H⁺ (aq) + Cl⁻ (aq) → Na⁺ (aq) + Cl⁻ (aq) + H₂O (l)

General Formulas and Concepts:

Atomic Structure

Compounds

Aqueous Solutions

Solubility Rules
States of matter

Stoichiometry

Reaction RxN Prediction
Balancing Reactions RxN

Explanation:

Step 1: Define

NaOH reacting w/ HCl

NaOH is soluble

HCl is soluble

[RxN] NaOH (aq) + HCl (aq) → NaCl (aq) + H₂O (l)

Step 2: Total Ionic Equation

Break up soluble compounds into ionic form.

[T.I.E] Na⁺ (aq) + OH⁺ (aq) + H⁺ (aq) + Cl⁻ (aq) → Na⁺ (aq) + Cl⁻ (aq) + H₂O (l)

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3 years ago

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3 years ago

What is the mass of 5 mole of ammonia . Calculate the number of NH₃ molecules, nitrogen atom and hydrogen atoms in it..

Aloiza [94]

Molar mass of NH_3

$\\ \sf\longmapsto 14u+3(1u)$

$\\ \sf\longmapsto 14u+3u$

$\\ \sf\longmapsto 17g/mol$

We know.

No of moles=Given mass/Molar mass

$\\ \sf\longmapsto Given\;Mass=17(5)$

$\\ \sf\longmapsto Given \:Mass\:of\:NH_3=85g$

Now

Lets write the balanced equation

$\\ \sf\longmapsto N_2+3H_2=2NH_3$

There is 2moles of Ammonia
3moles of H_2
1mole of N_2

Now

$\boxed{\sf No\:of\:Molecules =No\:of\;moles\times Avagadro\:no}$

For Hydrogen

$\\ \sf\longmapsto 3\times 6.023\times 10^{23}$

$\\ \sf\longmapsto 18.069\times 10^{23}$

$\\ \sf\longmapsto 1.8\times 10^{22}molecules$

For Ammonia

$\\ \sf\longmapsto 2\times 6.023\times 10^{23}$

$\\ \sf\longmapsto 12.046\times 10^{23}$

$\\ \sf\longmapsto 1.2\times 10^{22}molecules$

For Nitrogen

$\\ \sf\longmapsto 1\times 6.023\times 10^{23}$

$\\ \sf\longmapsto 6.023\times 10^{23}molecules$

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3 years ago