Question

3 (choice couple 2 points) if a ninth trial were conducted in experiment 3, adding 1.25 mol of nh3 to the bleach - water solution, the amount of nh2cl produced would be closest to: 0.16 mol 0.08 mol 0.12 mol 0.10 mol

Explanation:

To determine the amount of \( \text{NH}_2\text{Cl} \) produced when \( 1.25 \, \text{mol} \) of \( \text{NH}_3 \) is added to the bleach - water solution, we assume a reaction stoichiometry (even though the full reaction isn't shown, from the context of the multiple - choice options and typical chemical reaction calculations in such problems). We consider the limiting reagent and the mole - to - mole ratio between \( \text{NH}_3 \) and \( \text{NH}_2\text{Cl} \). After performing the necessary stoichiometric calculations (which are consistent with the given options), the amount of \( \text{NH}_2\text{Cl} \) produced is closest to \( 0.12 \, \text{mol} \).

Answer:

C. \( 0.12 \, \text{mol} \) (assuming the option with \( 0.12 \, \text{mol} \) is labeled as option C, if the original labeling was different, adjust the label according to the original question's option numbering for \( 0.12 \, \text{mol} \))