Stuff with limting reagents and products etc..help!?
2 Al(OH)3 (s)+ 3 H2SO4 (aq) --> Al2(SO4)3 (aq) + 6 H20(l)
which reagent is the limiting reagent when 0.500mol Al(OH)3 and 0.500mol H2SO4 are allowed to react?
how many moles of Al2(SO4)3 can form under these conditions?
how many moles of the excess reactant remain after the completion of the reaction?
could you please show some of your thought process when you answer this to makes sure i understand the whole processs? thanks
Answer:
The ratio between Al(OH)3 and H2SO4 is 2 : 3
because the number of moles is the same H2SO4 is the limiting reactant
The ratio between H2SO4 ( limiting reactant ) and Al2(SO4)3 is 3 : 1
3 : 1 = 0.500 : x
x = 0.167 moles of Al2(SO4)3 formed
the excess reactant is Al(OH)3
2 ( moles Al(OH)3) : 3 moles H2SO4 = x : 0.500
x = 0.333 moles Al(OH)3 needed for the reaction
0.500 - 0.333 = 0.167 moles Al(OH)3 in excess
I hope this help you
http://www.shodor.org/unchem/basic/stoic...
1. Since the balanced equation tells us you need 2 moles of Al(OH)3 for every 3 moles of H2SO4, then to react completely 0.5M of Al(OH)3 requires 0.75M of H2SO4 (which is more than there is available). 0.5M of H2SO4 will only react with 0.333M of Al(OH)3. The H2SO4 is therefore the limiting reagent and there will be 0.167M Al(OH)3 left unreacted.
2. Since for every 2 moles of Al(OH)3 you get 1 mole of Al2(SO4)3, it therefore follows that from 0.5 moles of Al(OH)3 you will get 0.25 moles of Al2(SO4)3.
3. See above - 0.167M Al(OH)3.
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