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3 years ago

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Chemistry

1 answer:

svp [43]3 years ago

3 0

Answer:

B is the answer

Explanation:

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Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and

rusak2 [61]

Answer:

$4Zn_(_s_)~+~7OH^-~_(_a_q_)~+~NO_3^-_(_a_q_)~+~6H_2O_(_l_)~-->4Zn(OH)_4^-^2_(_a_q_)~+~NH_3_(_g_)$

-) Oxidizing agent: $NO_3^-_(_a_q_)$

-) Reducing agent: $Zn_(_s_)$

Explanation:

The first step is separate the reaction into the semireactions:

A. $Zn~->Zn(OH)_4^-^2$

B. $NO_3^-~->~NH_3$

If we want to balance in basic medium we have to follow the rules:

1. We adjust the oxygen with $OH^-$

2. We adjust the H with $H_2O$

3. We adjust the charge with $e^-$

Lets balance the first semireaction A. :

$Zn~+~4OH^-~->Zn(OH)_4^-^2~+~2e^-$

Now, lets balance semireaction B:

$NO_3^-~+~8e^-~+~6H_2O~->~NH_3~+~9OH^-$

Finally, we have to add the two semireactions:

_________________________________________

$8~(Zn~+~4OH^-~->Zn(OH)_4^-^2~+~2e^-)$

$2~(NO_3^-~+~8e^-~+~6H_2O~->~NH_3~+~9OH^-)$

_________________________________________

$(8Zn~+~32OH^-~->8Zn(OH)_4^-^2~+~16e^-)$

$(2NO_3^-~+~16e^-~+~12H_2O~->~2NH_3~+~18OH^-)$

Cancel out the species on both sides:

$8Zn~+~14OH^-~+~2NO_3^-~+~12H_2O~-->8Zn(OH)_4^-^2~+~2NH_3$

Simplifying the equation :

$4Zn~+~7OH^-~+~NO_3^-~+~6H_2O~-->4Zn(OH)_4^-^2~+~NH_3$

The $Zn_(_s_)$ is oxidized therefefore is the reducing agent. The $NO_3^-_(_a_q_)$ is reduced therefore is the oxidizing agent.

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Answer:

123.0*12.35/(0.05*6.049)

first do in small bracket 123.0*12.35/0.30245

and then divide 123.0*40.83319557

and multiply 5022.483055

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