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2 years ago

What is the reducing agent in the following reaction? (PLEASE RESPOND FAST)

Chemistry

2 answers:

ahrayia [7]2 years ago

8 0

Answer:

A) Na

Explanation:

We analyze the change of electrons in the equation.

So:

S (0) --> S (-2)

The oxidizing agent is the one that gains electrons. Sulfur goes from 0 to -2 then it is gaining electrons.

Na (0) --> Na (+1)

The reducing agent is the one that loses electrons. Sodium goes from 0 to +1 which means that it loses electrons.

Therefore the reducing agent is Na.

stich3 [128]2 years ago

3 0

S is the reducing agent in this rxn

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Answer:

$\% Fe^{+2}=70%$

Explanation:

Hello,

In this case, we could considering this as a redox titration:

$Fe^{+2}+(Cr_2O_7)^{-2}\rightarrow Fe^{+3}+Cr^{+3}$

Thus, the balance turns out (by adding both hydrogen ions and water):

$Fe^{+2}\rightarrow Fe^{+3}+1e^-\\(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow 2Cr^{+3}+7H_2O\\\\6Fe^{+2}+(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow Cr^{+3}+7H_2O+6Fe^{+3}+6e^-$

Thus, by stoichiometry, the grams of Fe+2 ions result:

$m_{Fe^{+2}}=0.04021\frac{molK_2Cr_2O_7}{L}*0.02872L*\frac{6molFe^{+2}}{1molK_2Cr_2O_7}*\frac{56gFe^{+2}}{1molFe^{+2}}=0.388gFe^{+2}$

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Best regards.

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2 years ago