Asked by Em
When doing a limiting reagant problem, I need to determine to mole ratio, right? But the problem is if 19.6 g of H3PO4 reacted with 22.4 g KOH...
How do I get two separate mole ratios?
The answers are: 3/1 for H3PO4 and 3/3 for KOH, but why?
How do I get two separate mole ratios?
The answers are: 3/1 for H3PO4 and 3/3 for KOH, but why?
Answers
Answered by
bobpursley
You balance the reaction equation.
H3PO4+3KOH>> K3PO4 + 3HOH
so you need 3 moles KOH for each mole phosporic acid.
figure the moles you have of each:
KOH=22.4g/56.1= about .4 you do it.
H3PO4=19.6/97about=about .5
so clearly, you do not have nearly enough KOH(you would need almost four times as muchas you have (.5*3/.4).
So the KOH is going to determine how much product you make.
H3PO4+3KOH>> K3PO4 + 3HOH
so you need 3 moles KOH for each mole phosporic acid.
figure the moles you have of each:
KOH=22.4g/56.1= about .4 you do it.
H3PO4=19.6/97about=about .5
so clearly, you do not have nearly enough KOH(you would need almost four times as muchas you have (.5*3/.4).
So the KOH is going to determine how much product you make.
Answered by
Em
Thank you.
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