The reaction that takes place during charging of the lead storage cell is $$2\text{PbSO}_4(s) + 2\text{H}_2\text{O}(l) \rightarrow \text{Pb}(s) + \text{PbO}_2(s) + 2\text{H}_2\text{SO}_4(aq)$$. If a current of 10.0 A is passed for 1.50 h for charging, the amount of $$\text{PbSO}_4$$ reacted is

A

25.0 g

B

56.0 g

C

120.5 g

D

170.0 g