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: 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 [/sub]
