Solution Stoichiometry

 

72) m1 = 7.52 g Sr(NO3)2          V2 = 0.100 L          [Na2CrO4] = 0.0425 M

      V1 = 0.750 L

      

      Sr(NO3)2(aq) + Na2CrO4(aq) 2NaNO3(aq) + SrCrO4(s)

      Sr2+(aq) + CrO42-(aq) SrCrO4(s)

 

      [Sr(NO3)2] = n/V = 7.52 g Sr(NO3)2 x 1 mol Sr(NO3)2/211.64 g Sr(NO3)2/0.750 L = 0.0474 M

      nSr2+ = 0.0474 mol Sr(NO3)2/L x 0.100 L x 1 mol Sr2+/1 mol  Sr(NO3)2 = 4.74 x 10-3 mol Sr2+   

      nNa2CrO4 = 4.74 x 10-3 mol Sr2+ x 1 mol CrO42-/1 mol Sr2+ x 1 mol Na2CrO4/1 mol CrO42- = 4.74 x 10-3 mol Na2CrO4

 

      [Na2CrO4] = n/V

      V = n/[Na2CrO4] = 4.74 x 10-3 mol Na2CrO4/0.0425 M Na2CrO4 = 0.112 L Na2CrO4