# chemistry

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!!!

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1. 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.

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posted by DrBob222

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