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]
- 0.8g
-
4.0g
- 40.0g
- 80.0g
The correct answer is: B
Explanation
Equation of the reaction
H\(_2\)SO\(_4\)(g) + CuO(s) → CuSO\(_4\)(aq) + H\(_2\)O
Relative molecular mass of H\(_2\)SO\(_4\) = 1 + 2 + 32 + 16 + 4 = 98
Relative molecular mass of CuO = 64 + 16 =80
From the above equation, 98g of H\(_2\)SO\(_4\) reacts with 80g of CuO.
∴ 4.9g of H\(_2\)SO\(_4\) will react with Xg of CuO
= (4.9 * 80)/98 of CuO = 392/98 = 4.0g
Alternatively,
H\(_2\)SO\(_4\)(g) + CuO(s) → CuSO\(_4\)(aq) + H\(_2\)O
1 : 1
Since the combining ratio of H\(_2\)SO\(_4\) and CuO is in ratio 1:1, the number of moles will also be equal.
Hence, we can first calculate the amount of H\(_2\)SO\(_4\), since we were given the mass of H\(_2\)SO\(_4\) that reacted as 4g and the Molecular mass is 98g/mol
Amount of moles of H\(_2\)SO\(_4\) = \(\frac{mass}{Molecular mass}\)
= \(\frac{4.9}{98}\) = 0.05 mole
Amount of moles of CuO = \(\frac{mass}{Molecular mass}\)
0.05 = \(\frac{mass}{64 + 16}\)
Mass = 0.05 x 80 = 4.0g