\(Zn + H_2 SO_4 → ZnSO_4 + H_2\)
in the above reaction, how much zinc will be left undissolved if 2.00g of zinc is treated with 10cm^3 of 1.0 M of \(H_2SO_4\)?
[Zn =65, s = 32, O = 16, H = 1]
-
1.35g
- 1.00g
- 0.70g
- 0.65g
- 0.00g
The correct answer is: A
Explanation
\(Zn + H_2 SO_4 → ZnSO_4 + H_2\)
1 mole of Zn = 1 mole \(H_2SO_4\)
no of moles of \(H_2SO_4\) = C X V = 1 X 10/1000 = 0.01moles.
0.01mole \(H_2SO_4\) = 0.01 mole Zn
mass of Zn = 0.01 x 65(given) = 0.65g
therefore, the mass of undissolved Zn = 2.00(given) - 0.65 = 1.35g.