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If a solution contains 4.9g of tetraoxosulphate (VI) acid...

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]

A.

0.8g

B.

4.0g

C.

40.0g

D.

80.0g

Correct answer is B

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