if 31.2 mL of 0.45 M sodium hydroxide is mixed with 65.4 mL of 0.088 M phosphoric acid, how many kJ of heat are produced?

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  • 3 weeks ago

    3 NaOH + H3PO4 → 3 H2O + Na3PO4

    OH{-}(aq) + H{+}(aq) → H2O(l), ΔH = -57.5 kJ

    (31.2 mL) x (0.45 M NaOH) = 14.04 mmol NaOH

    (65.4 mL) x (0.088 M H3PO4) = 5.7552 mmol H3PO4

    14.04 millimoles of NaOH would react completely with 14.04 x (1/3) = 4.68 millimoles of H3PO4, but there is more H3PO4 present than that, so H3PO4 is in excess and NaOH is the limiting reactant.

    (14.04 × 10^-3 mol NaOH) x (57.5 kJ/mol) = 807.3 J = 800 J

  • 3 weeks ago

    Its is the 69 baby

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