What mass (g) of agbr is formed when 41.8 ml of 0.152 m agno3 is treated with an excess of aqueous hydrobromic acid?

m AgBr = 1.193 g

Explanation:

∴ V AgNO3 = 41.8 mL = 0.0418 L

∴ [ AgNO3 ] = 0.152 mol/L

∴ Mw AgBr = 187.77 g/mol

⇒ n AgNO3 = (0.0418 L)(0.152 mol/L) = 6.3536 E-3 mol AgNO3

⇒ n AgBr = (6.3536 E-3 mol AgNO3)(mol AgBr/mol AgNO3) = 6.3536 E-3 mol AgBr

⇒ m AgBr =  (6.3536 E-3 mol AgBr)(187.77 g/mol) = 1.193 g AgBr

the right answer is c i just did the question.