Chemistry question?
2017-10-25 10:01 am
When 100.0 mL of 2.0 M Al(NO3)3 is added to 200.0 mL of 2.0 M KOH, how many moles of Al(OH)3 can be produced?
回答 (1)
2017-10-25 10:09 am
Initial number of moles of Al(NO₃)₃ = (2.0 mol/L) × (100.0/1000 L) = 0.2 mol
Initial number of moles of KOH = (2.0 mol/L) × (200.0/1000 L) = 0.4 mol
Balanced equation for the reaction :
Al(NO₃)₃ + 3KOH → Al(OH)₃ + 3KNO₃
Mole ratio Al(NO₃)₃ : KOH = 1 : 3
When KOH completely reacts, Al(NO₃)₃ needed = (0.4 mol) × (1/3) = 0.133 mol < 0.2 mol
Hence, Al(NO₃)₃ is in excess, and KOH is the limiting reactant/reagent.
According to the above equation, mole ratio KOH : Al(OH)₃ = 3 : 1
Number of moles of Al(OH)₃ produced = (0.4 mol) × (1/3) = 0.133 mol
Initial number of moles of KOH = (2.0 mol/L) × (200.0/1000 L) = 0.4 mol
Balanced equation for the reaction :
Al(NO₃)₃ + 3KOH → Al(OH)₃ + 3KNO₃
Mole ratio Al(NO₃)₃ : KOH = 1 : 3
When KOH completely reacts, Al(NO₃)₃ needed = (0.4 mol) × (1/3) = 0.133 mol < 0.2 mol
Hence, Al(NO₃)₃ is in excess, and KOH is the limiting reactant/reagent.
According to the above equation, mole ratio KOH : Al(OH)₃ = 3 : 1
Number of moles of Al(OH)₃ produced = (0.4 mol) × (1/3) = 0.133 mol
收錄日期: 2021-05-01 14:09:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20171025020135AAnoib9