Asked by Colin
A solution of .12 L of 0.160 M KOH is mixed with a solution of .3 L of 0.230 M NiSO4.
the equation for this reaction is:
2KOH (aq)+NiSO4 (aq) ----->K2SO4 (aq)+Ni(OH)2 (s)
.89 grams of Ni(OH)2 precipitate form
I need to know the concentration remaining in the solution of:
1) Ni (II)
2) SO4
3) K
I do not know how to do these nor can i find aid to show me how to do it step by step, so to whoever answers this question, could you also show steps so i can learn how to do this? thank you to whoever puts in their time and helps me!!
the equation for this reaction is:
2KOH (aq)+NiSO4 (aq) ----->K2SO4 (aq)+Ni(OH)2 (s)
.89 grams of Ni(OH)2 precipitate form
I need to know the concentration remaining in the solution of:
1) Ni (II)
2) SO4
3) K
I do not know how to do these nor can i find aid to show me how to do it step by step, so to whoever answers this question, could you also show steps so i can learn how to do this? thank you to whoever puts in their time and helps me!!
Answers
Answered by
DrBob222
The first part of this is a limiting reagent problem. I solve these problems by solving two stoichiometry problems. The first one using reagent 1 and all of the other needed; the second time with reagent 2 and all of the other needed. You will obtain two answers; obviously both can't be right. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Here is a step by step procedure for solving stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html
http://www.jiskha.com/science/chemistry/stoichiometry.html
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