A solution is made by mixing 15.0g of Sr(OH)2 and 35.0mL of 0.250 M HNO3. Calcul

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Sr(OH)2 +2 HNO3 >> Sr(NO3)2 + 2H2O Anyway this is what I do. Calculate the moles of each reactant you have. For Sr(OH)2 use 15.0 g / 121.64 g/mol = 0.1233 mol For HNO3 it is C x V = 0.250 mol/L x 0.035 L = 0.00875 Now look at the mol ratio between Sr(OH)2 and HNO3. Can yo see that it is 1 : 2? So for every mol of Sr(OH)2 you need TWICE as many mol of HNO3. Well do you have that many mol of HNO3? Well you have 0.1233 mol of Sr(OH)2, so twice as much would make it 0.1233 x 2 or 0.2466mol Well you only have 0.00875 mol of HNO3. Clearly not enough to use up all the Sr(OH)2. So HNO3 is the limiting reagent. The reaction will stop when 0.00875 mol of HNO3 is used up Now how many moles of Sr(OH)2 will be used up? Well work in reverse. You only have 0.000875 mol of HNO3 but you only need half that amount for the Sr(OH)2. So 0.00875 / 2 = 0.004375 mol of Sr(OH)2 will be used. Now all you have to do is find the concentration of that much Sr(OH)2. C = n / V so 0.004375 mol / 0.035 mL = 0.125 mol/L That is the concentration of Sr(OH)2 remaining. Since Sr(OH)2 --------------> Sr+2(aq) + 2OH-(aq), then the concentration of Sr+2 = 0.115 mol/L and the conc of OH- will be 2 x 0.125 mol/L or 0.250 mol/L

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