Question

When carboxylic acid is mixed with aqueous ammonia solution, ammonium carbonate is formed:

H2CO3 (aq) + 2 NH3 (aq) → (NH4)2CO3(aq)

a. If 10.00 mL of a 0.105 M NH3 solution reacted with 25.00 mL of a 0.0835 M H2CO3 solution, which
reactant is the limiting reagent? (6 pts)
b. What is the maximum number of moles of ammonium carbonate that can be formed? (3 pts)
c. What is the molarity of the ammonium carbonate solution given that the two reagent solution volumes
are additive? (3 pts)
d. If 0.423 moles of ammonium carbonate actually formed, what is the theoretical yield? (3 pts)

Answer 1

HUN the COB ago + 2NH3 cays (NH4), CO3(aq) ® 10,00 mL of 0.105 M N N =) 1.05 umol NH₂ 25.00 mL of 0,0835 M H coz a) 2,0875 mmol hcom As per clio chiometry (mot the com 2 mol Nih 4.175" So, 2.0875 mmol the loz - Nithy but we have only 105 munol so Nithy is uniting reagent since it is reagent, basis of reaction will be the dos 2 mot With Imor (NH4), com So, in 1.05 mmol Nth 1 x 1,05 mmol (NH) 23 1 0.525 mnol. ſ

ЛЧ mare, so mare, Ammonium number of carbonate moles. fermed وما 0 | 1 - are 5. 25 x 10 volume a 10.00 mL + 25.00mL - Total = 35.00 ml 5.25 80 holl moty 1 - morarity molanti & 35.00 x103 (4) Volume volume of ammonium carbonate in molanty 0.015 M & wo yield = Actual yield xlo Theonticel Yield ^ Oesty 0.423 x lov 0,524 7. yield – 80.7%.