Question

A.

Calculate the product molar concentration in 50 mL of total reaction solution in the following reaction and conditions. Presume that all of the SCN– ions react.     

please show calculations

Fe3+ (aq) + SCN– (aq) → FeSCN2+ (aq)

Beaker	0.0020 M SCN– (mL)	[FeSCN2+] (M)
A	1.0
B	2.0
C	3.0
D	4.0

B.

Calculate the initial concentrations of the Fe3+ and the SCN– ions in the following reaction:

please show calculations

Test Tube	0.0020 M Fe3+ (mL)	0.0020 M SCN– (mL)	H2O (mL)	[Fe3+] (M)	[SCN-] (M)
A	3.00	3.00	4.00
B	3.00	4.00	3.00
C	3.00	5.00	2.00

Answer 1

A. For each Fe³⁺ one SCN^- is reacting,

A. V₁ = 1 ml; M₁ = 0.002M; V₂ = 50 ml; M₂ = ?

V₁ M₁ = V₂ M₂

M₂ = V₁ M₁ / V₂ = 1 x 0.002 / 50 = 4 x 10^-5 M

B. V₁ = 2 ml; M₁ = 0.002M; V₂ = 50 ml; M₂ = ?

V₁ M₁ = V₂ M₂

M₂ = V₁ M₁ / V₂ = 2 x 0.002 / 50 = 8 x 10^-5 M

C. V₁ = 3 ml; M₁ = 0.002M; V₂ = 50 ml; M₂ = ?

V₁ M₁ = V₂ M₂

M₂ = V₁ M₁ / V₂ = 3 x 0.002 / 50 = 1.2 x 10^-4 M

D. V₁ = 4 ml; M₁ = 0.002M; V₂ = 50 ml; M₂ = ?

V₁ M₁ = V₂ M₂

M₂ = V₁ M₁ / V₂ = 4 x 0.002 / 50 = 1.6 x 10^-4 M

B. Initial concentration of both the species are given in the table itself. If you take that solution 3 or 4 or 5 ml the concentration won't change at all. I don't think your question is clear. Please check it and comment in my answer, then I will be able to complete it. Thanks