Question

A solution is made by mixing 15.0 g of sr(oh)2 and 55.0 ml of 0.200 m hno3.

Answer 1

 Letter A: The balanced equation is:

Sr(OH)2 (s) + 2 HNO3 (aq) → Sr(NO3)2 (aq) + 2 H2O 

Letter B: The solution is:

(15.0 g Sr(OH)2) / (121.6358 g Sr(OH)2/mol) = 0.12332 mol Sr(OH)2 
(0.0550 L) x (0.200 mol/L HNO3) = 0.011 mol HNO3 

0.011 mole of HNO3 would respond entirely with 0.011 x (1/2) = 0.0055 mole of Sr(OH)2 but there is more Sr(OH)2 existing than that, so Sr(OH)2 is in excess and HNO3 is the limiting reactant. 


( 0.011 mol HNO3) x (1/2) = 0.0055 mol Sr(NO3)2 


Since the NO3{-} ions continued in solution during, the concentration of the NO3{-} ions didn't change during the reaction, so they are still 0.200 M. 
The concentration of Sr{+2} ions is (0.0055 mol Sr(NO3)2) / (55.0 mL) = 0.100 M 
Since HNO3 is the limiting reactant, no H{+} ions continue at the end of the reaction. 

Letter C: Since all of the acid reacted and there was still Sr(OH)2 left over, the resulting solution would be basic.