MCQ
From the following $0.1\, M$ solution of compound, will be basic in nature
- A$NH_4Cl$
- ✓$CH_3COONa$
- C$Na_2SO_4$
- D$CH_3COONH_4$
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$20 \;g \;\;\;\;5 \;g$
Consider the above reaction, the limiting reagent of the reaction and number of moles of $NH _{3}$ formed respectively are