# When 0.121 moles of HCl reacts with 0.399 moles of NH3, how much NH3 is consumed?

## NH3+HCl -----> NH4Cl

Approximately $0.121 m o l$ $N {H}_{3}$.
$N {H}_{3} + H C l \to N {H}_{4} C l$
Fairly simple question. Consider the moles and coefficients of the reaction, and do some simple addition and subtraction to "get" the amount of $N {H}_{3}$ consumed.