Use O = 16 and molar volume 22.4 L/mol at STP. Ammonia reduces hot copper(II) oxide: 2NH3 + 3CuO -> N2 + 3Cu + 3H2O. When 4.48 L of NH3 (at STP) is passed completely over excess CuO, the loss in mass of the solid in the tube is:

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Use O = 16 and molar volume $22.4 L/mol$ at STP. Ammonia reduces hot copper(II) oxide: $2 NH X_{3} + 3 CuO N X_{2} + 3 Cu + 3 H X_{2} O$ . When $4.48 L$ of $NH X_{3}$ (at STP) is passed completely over excess $CuO$ , the loss in mass of the solid in the tube is:

n (NH X_{3}) = 4.48/22.4 = 0.2

CuO

reacted

= 0.2 \times 3/2 = 0.3

mol, each losing one O atom to the gas phase as water. Mass lost

= 0.3 \times 16 = 4.8 g

(the solid becomes Cu).