Question

In: Chemistry

in a test tube containing 2 mL of 2.4*10^-4 M KSCN, and 23 mL of 0.10...

in a test tube containing 2 mL of 2.4*10^-4 M KSCN, and 23 mL of 0.10 M Fe(NO3)3, what are the number of moles of SCN^-, the number of moles of FeSCN^2+, and the concentration of FeSCN^2+ in mol/L?

Expert Solution

2 mL of 2.4*10^-4 M KSCN,

no of moles of KSCN = molarity * volume in L

= 2.4*10^-4 *0.002

= 4.8*10^-7 moles

KSCN -----> K⁺ + SCN^-

4..8*10^-7 moles 4.8*10^-7 moles

23 mL of 0.10 M Fe(NO3)3

no of moles of Fe(NO3)3 = molarity *volume in L

= 0.1*0.023 =0.0023 moles

Fe(NO3)3 ------> Fe⁺³ + 3NO3^-

0.0023 moles 0.0023 moles

Fe⁺³ + SCN^- -----> FeSCN^2+

SCN^- is limiting reageent

1 mole of Fe³⁺ react with SCN^- to form 1 mole of FeSCN^2+

4.8*10^-7 moles of Fe³⁺ react with SCN^- to form 4.8*10^-7 moles of FeSCN^2+

total volume is = 2+23 = 25 ml = 0.025 L

molarity of FeSCN^2+ = no of moles/ volume in L

= 4.8*10^-7/0.025 = 1.92*10^-5 M

queen_honey_blossom answered 2 years ago

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