Asked by Sara
how many mL or .0252M HNO3 are required to neutralize 25.00mL of .0148M of Sr(OH)2?
and how did you get the answer!!!
and how did you get the answer!!!
Answers
Answered by
DrBob222
Sr(OH)2 + 2HNO3 ==> Sr(NO3)2 + 2H2O
1. Write and balance the equation as above.
2. How many mols do you have to neutralize? That's mols Sr(OH)2 = M x L = ?
3. Using the coefficients in the balanced equation, convert mols Sr(OH)2 to mols of the other reagent. In this case,
mols HNO3 = mols Sr(OH)2 x (2 mols HNO3/1 mol Sr(OH)2) = mols Sr(OH)2 x (2/1) = ?
This four-step procedure will work almost all of the titration problems you have.
4. Now convert mols HNO3 to volume using M = mols/L. You know M and mols, solve for L and convert to mL.
1. Write and balance the equation as above.
2. How many mols do you have to neutralize? That's mols Sr(OH)2 = M x L = ?
3. Using the coefficients in the balanced equation, convert mols Sr(OH)2 to mols of the other reagent. In this case,
mols HNO3 = mols Sr(OH)2 x (2 mols HNO3/1 mol Sr(OH)2) = mols Sr(OH)2 x (2/1) = ?
This four-step procedure will work almost all of the titration problems you have.
4. Now convert mols HNO3 to volume using M = mols/L. You know M and mols, solve for L and convert to mL.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.