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If a solution contains 4.9g of tetraoxosulphate (VI) acid, calculate the amount ...


Question

If a solution contains 4.9g of tetraoxosulphate (VI) acid, calculate the amount of copper (II) oxide that will react with it

[Cu = 64, O = 16, S = 32, H = 1]

Options

A) 0.8g

B) 4.0g

C) 40.0g

D) 80.0g


The correct answer is B.

Explanation:

Equation of the reaction
H2SO4(g) + CuO(s) → Cu4(aq) + H2O
Relative molecular mass of H2SO4 = 1 + 2 + 32 + 16 + 4 = 98
Relative molecular mass of CuO = 64 + 16 =80
From the above equation, 98s of H2SO4 reacts with 80s of CuO.
∴ 4.9g of H2SO4 will react with
= (4.9 * 80)/98 of CuO = 392/98 = 4.0g

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Dicussion (1)

  • Equation of the reaction
    H2SO4(g) + CuO(s) → Cu4(aq) + H2O
    Relative molecular mass of H2SO4 = 1 + 2 + 32 + 16 + 4 = 98
    Relative molecular mass of CuO = 64 + 16 =80
    From the above equation, 98s of H2SO4 reacts with 80s of CuO.
    ∴ 4.9g of H2SO4 will react with
    = (4.9 * 80)/98 of CuO = 392/98 = 4.0g

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