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  • one year ago

The volume of sodium hydroxide solution used was 675.0 mL, and the concentration was 0.875 M. Calculate the moles of sulfuric acid that were neutralized (assuming that the reaction goes to completion). 2NaOH (aq) + H2SO4 (aq) → Na2SO4 (aq) + 2H2O (l)

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  1. aaronq
    • one year ago
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    1. First find the moles of NaOH used with the volume and molarity given. \(\sf molarity=\dfrac{moles~of~solute}{L_{solution}}\) 2. Next use the coefficient and the moles in a ratio: \(\sf \dfrac{moles~of~NaOH}{NaOH's ~coefficient}=\dfrac{moles~of~H_2SO_4}{H_2SO_4's~coefficient}\) Plug in moles of NaOH and both coefficients, solve for moles of H2SO4

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