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) + CrO₄^2-(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

      n_Sr2+ = 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⁺   

      n_Na2CrO4 = 4.74 x 10^-3 mol Sr²⁺ x 1 mol CrO₄^2-/1 mol Sr²⁺ x 1 mol Na₂CrO₄/1 mol CrO₄^2- = 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