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3 years ago

Writing the net ionic equation

Chemistry

2 answers:

olganol [36]3 years ago

7 0

Answer: The net ionic equation is written below.

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of silver nitrate and ammonium sulfide is given as:

$2AgNO_3(aq.)+(NH_4)_2S(aq.)\rightarrow Ag_2S(s)+2NH_4NO_3(aq.)$

Ionic form of the above equation follows:

$2Ag^+(aq.)+2NO_3^-(aq.)+2NH_4^+(aq.)+S^{2-}(aq.)\rightarrow Ag_2S(s)+2NH_4^+(aq.)+2NO_3^-(aq.)$

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$2Ag^+(aq.)+S^{2-}(aq.)\rightarrow Ag_2S(s)$

Hence, the net ionic equation is written above.

vazorg [7]3 years ago

3 0

First write the molecular equation with states:

(NH4)2S (aq) + 2AgNO3(aq) → Ag2S (s) + 2NH4NO3

Now write a full ionic equation by separating into ions all substances that dissociate: anything (s) (g) or (l) does not dissociate

2NH4 + (aq) + S 2-(aq) + 2Ag+ (aq) + 2NO3- (aq) → Ag2S(s) + 2NH4 + (aq) + 2NO3- (aq)

To write the NET IONIC equation, inspect the full ionic equation above and delete anything that appears on both sides of the → sign:

Net ionic equation:

S 2-(aq) + 2Ag + (aq) → Ag2S(s)

You might be interested in

In the following reaction, how many grams of NaBr will produce 244 grams of NaNO3? Pb(NO3)2(aq) 2 NaBr(aq) PbBr2(s) 2 NaNO3(aq)

lyudmila [28]

Answer : The mass of $NaBr$ is, 295.323 grams

Solution :

First we have to calculate the moles of $NaNO_3$ .

$\text{Moles of }NaNO_3=\frac{\text{Mass of }NaNO_3}{\text{Molar mass of }NaNO_3}=\frac{244g}{85g/mole}=2.87moles$

Now we have to calculate the moles of NaBr.

The balanced chemical reaction is,

$Pb(NO_3)_2(aq)+2NaBr(aq)\rightarrow PbBr_2(s)+2NaNO_3(aq)$

From the balanced reaction we conclude that

As, 2 moles of $NaNO_3$ react with 2 moles of $NaBr$

So, 2.87 moles of $NaNO_3$ react with 2.87 moles of $NaBr$

Now we have to calculate the mass of NaBr.

$\text{Mass of }NaBr=\text{Moles of }NaBr\times \text{Molar mass of }NaBr$

$\text{Mass of }NaBr=(2.87mole)\times (102.9g/mole)=295.323g$

Therefore, the mass mass of $NaBr$ is, 295.323 grams

7 0

3 years ago

Use standard enthalpies of formation to calculate Δ H ∘ rxn for each reaction. MISSED THIS? Read Section 7.9; Watch KCV 7.9, IWE

Eva8 [605]

Answer:

Standard Heat of Reaction 1 = -136.2 kJ/mol

Standard Heat of Reaction 2 = -41.166 kJ/mol

Standard Heat of Reaction 3 = -136.07 kJ/mol

Standard Heat of Reaction 4 = 279.448kJ/mol

Explanation:

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

The required standard heat of formation for each of the reactants and product above, as obtained from literature is listed below.

C₂H₄ (g), 52.5 kJ/mol

H₂ (g), 0 kJ/mol

C₂H₆ (g), -83.7 kJ/mol

CO (g), -110.525 kJ/mol

H₂O (g), -241.818 kJ/mol

H₂ (g), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

NO₂ (g), 33.2 kJ/mol

H₂O (l), -285.8 kJ/mol

HNO₃ (aq), -206.28 kJ/mol

NO (g), 90.29 kJ/mol

Cr₂O₃ (s), -1128.4 kJ/mol

CO (g), -110.525 kJ/mol

Cr (s), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

Note that