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]
-
A.
0.8g -
B.
4.0g -
C.
40.0g -
D.
80.0g
Correct Answer: Option 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