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kifflom [539]
3 years ago
9

If you assume this reaction is driven to completion because of the large excess of one ion, what is the concentration of [Fe(SCN

)]2+ that would be formed from 6.00 mL of 0.00180 M KSCN 5.00 mL 0.240 M Fe(NO3)3 and 14.00 mL of 0.050 M HNO3.
Question 3 options:

0.240 M

4.32 x 10^-4

0.0480

0.0460 M
Chemistry
1 answer:
viktelen [127]3 years ago
7 0

Answer : The concentration of [Fe(SCN)]^{2+} is, 4.32\times 10^{-4}M

Explanation :

When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is SCN^- and Fe^{3+} is excess reagent.

First we have to calculate the moles of KSCN.

\text{Moles of }KSCN=\text{Concentration of }KSCN\times \text{Volume of solution}

\text{Moles of }KSCN=0.00180M\times 0.006L=1.08\times 10^{-5}mol

Moles of KSCN = Moles of K^+ = Moles of SCN^- = 1.08\times 10^{-5}mol

Now we have to calculate the concentration of [Fe(SCN)]^{2+}

\text{Concentration of }[Fe(SCN)]^{2+}=\frac{\text{Moles of }[Fe(SCN)]^{2+}}{\text{Volume of solution}}

Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L

\text{Concentration of }[Fe(SCN)]^{2+}=\frac{1.08\times 10^{-5}mol}{0.025L}=4.32\times 10^{-4}M

Thus, the concentration of [Fe(SCN)]^{2+} is, 4.32\times 10^{-4}M

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