1 mole H2SO4 = 98.078g H2SO4

Grams NaOH??

Balanced equation.

2NaOH + H2SO4 --> Na2SO4 + 2H2O

4.9 grams H2SO4 (1 mole H2SO4/98.086 grams)(2 mole NaOH/1 mole H2SO4)(39.998 grams/1 mole NaOH)

= 4.0 grams NaOH needed

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We need to go from grams of CuO to mL of H2SO4.

Atomic weight of CuO = 63.55 g Cu + 16 g O = 79.55 g CuO

(.80 g CuO) * (1 mol CuO / 79.55 g CuO) = .0100566 mol CuO (g CuO cancel)

Since the moles of CuO is a 1:1 ratio to H2SO4 (see balanced equation) we know that: mol CuO = mol H2SO4 or 0.0100566 mol CuO = 0.0100566 mol H2SO4

3.0 M of H2SO4 means that there is 3 mol / 1 L. So we can divide this by the moles to get L then mL of H2SO4

(0.0100566 mol H2SO4) * (1 L H2SO4 / 3 mol H2SO4) * (1000 mL H2SO4 / 1 L H2SO4) = 3.4 mL H2SO4 (mol H2SO4 and L H2SO4 cancel)

So 3.4 ml of H2SO4 is needed to react with 0.80 g of CuO.

1 gm mole of Sulphuric Acid (H2SO4) weights 98 gm.

what is the intermediate after adding H2SO4 on naphthol?