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]
[Cu = 64, O = 16, S = 32, H = 1]
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Correct Answer: Option B
Explanation:
Equation of the reaction
H2SO4(g) + CuO(s) → Cu4(aq) + H[sub]2O
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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Equation of the reaction
H2SO4(g) + CuO(s) → Cu4(aq) + H[sub]2O
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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