Question

What quantity in moles of Na₂S will be required to completely react with 49.0 mL of 0.200 M Fe(NO₃)₃?

A) 0.024 moles
B) 0.049 moles
C) 0.098 moles
D) 0.196 moles

Answer 1

Tthe moles of Na₂S required to react with Fe(NO₃)₃, stoichiometry is used to convert the volume and molarity of Fe(NO₃)₃ to moles and then apply the stoichiometric ratio from the balanced equation. There seems to be a discrepancy in the provided answer options as the calculated figure does not match any options.

Step-by-step explanation:

The question asks about the amount of Na₂S required to react completely with a given volume and molarity of Fe(NO₃)₃. To solve this problem, we need to use stoichiometry.

First, we find the number of moles of Fe(NO₃)₃ using the volume and molarity:

• Number of moles of Fe(NO₃)₃ = 49.0 mL x (1 L/1000 mL) x 0.200 M = 0.0098 moles

Assuming the balanced chemical equation is:

• Fe(NO₃)₃ + 3 Na₂S → Fe₂S₃ + 6 NaNO₃

We can see that it takes 3 moles of Na₂S to react with 1 mole of Fe(NO₃)₃. Therefore, the moles of Na₂S needed are:

• Moles of Na₂S = 3 x Number of moles of Fe(NO₃)₃
  = 3 x 0.0098 moles = 0.0294 moles

This result is not exactly one of the options provided, suggesting either a typo or miscalculation in the question. Without the proper balanced equation or correct options given, we cannot definitively select an answer from the options A through D. However, the calculation process demonstrates how to approach this kind of stoichiometry problem.