Question

lei a-131 chemistry

1. calculate the mass of fe(scn)₃ produced when 29.1 g of kscn reacts with excess fecl₃. show all work and include units.

Explanation:

Step1: Write the balanced reaction

The reaction between \( \text{KSCN} \) and \( \text{FeCl}_3 \) is: \( \text{FeCl}_3 + 3\text{KSCN}
ightarrow \text{Fe(SCN)}_3 + 3\text{KCl} \)

Step2: Calculate moles of \( \text{KSCN} \)

Molar mass of \( \text{KSCN} \) (K: 39.10 g/mol, S: 32.07 g/mol, C: 12.01 g/mol, N: 14.01 g/mol) is \( 39.10 + 32.07 + 12.01 + 14.01 = 97.19 \, \text{g/mol} \)
Moles of \( \text{KSCN} = \frac{\text{mass}}{\text{molar mass}} = \frac{29.1 \, \text{g}}{97.19 \, \text{g/mol}} \approx 0.2994 \, \text{mol} \)

Step3: Relate moles of \( \text{KSCN} \) to \( \text{Fe(SCN)}_3 \)

From the balanced equation, 3 moles of \( \text{KSCN} \) produce 1 mole of \( \text{Fe(SCN)}_3 \)
Moles of \( \text{Fe(SCN)}_3 = \frac{0.2994 \, \text{mol} \, \text{KSCN}}{3} \approx 0.0998 \, \text{mol} \)

Step4: Calculate molar mass of \( \text{Fe(SCN)}_3 \)

Fe: 55.85 g/mol, S: 32.07 g/mol (3 S), C: 12.01 g/mol (3 C), N: 14.01 g/mol (3 N)
Molar mass = \( 55.85 + 3\times(32.07 + 12.01 + 14.01) = 55.85 + 3\times58.09 = 55.85 + 174.27 = 230.12 \, \text{g/mol} \)

Step5: Calculate mass of \( \text{Fe(SCN)}_3 \)

Mass = moles × molar mass = \( 0.0998 \, \text{mol} \times 230.12 \, \text{g/mol} \approx 22.97 \, \text{g} \)

Answer:

The mass of \( \text{Fe(SCN)}_3 \) produced is approximately \( 23.0 \, \text{g} \) (or more precisely \( \approx 23.0 \, \text{g} \))