[hupnosis]

What mass of AgBr is formed when 35.5 mL of 0.184 M AgNO3 is treated with an excess of hydrobromic acid?

I was wondering if i did this problem correctly.
.0035*.184M=.006532 moles AgNO3
.006532 moles AgNO3 * 169.845g AgNO3 = 1.11g AgNO3
(1.11g AgNO3)(1 mole AgNO3/169.845)(1 mole AgBr/1 mole AgNO3)(187.768AgBr/1 mole AgBr)= 1.227g AgBr formed

so 1.227g of AgBr are formed.
Thanks for all the help

[mikasaur]

Besides the extra step, it looks good (you multiplied the 0.006532 mols of AgNO₃ by its atomic mass just to divide by that same mass in your next step).

Also you're using 0.0035 L instead of 0.00355 L.

I'd double check your atomic masses as well (not that they're wrong, but I didn't check them and it's always good to make sure you get easy stuff like that right!). And your sig figs!