Chemistry, 20.03.2020 11:14 ciarrap552
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
Answers: 2
![[Fe(SCN)]^{2+}](/tpl/images/0556/0307/0c409.png)
is, 
and 
is excess reagent.
= Moles of 
![\text{Concentration of }[Fe(SCN)]^{2+}=\frac{\text{Moles of }[Fe(SCN)]^{2+}}{\text{Volume of solution}}](/tpl/images/0556/0307/e16b4.png)
![\text{Concentration of }[Fe(SCN)]^{2+}=\frac{1.08\times 10^{-5}mol}{0.025L}=4.32\times 10^{-4}M](/tpl/images/0556/0307/cd669.png)
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