You mix 60. mL of 1.00 M sliver nitrate with 25 mL of 1.30 M sodium chloride. What mass of silver chloride ...?

First find limiting reagant, which will be the sodium chloride. The number of moles will be 0.0325 moles, then you multiply that number by molar mass of silver chloride (143.35 g/mol), and you get 4.65g. So I think A) should be the answer. Just in case, check yourself.