[winx]

we got this question as a bonus on a test and the teacher isnt taking up the answer i was wondering if anyone could help me with how to get started? This is the question:
0.790 g of silver nitrate and 0.473 g of potassium bromate are added to 379 mL water. Solid silver bromate is formed, dried, and weighed. What is the mass, in g, of the precipitated silver bromate? (Be careful to enter the correct number of significant figures. Do not enter units.) Assume silver bromate is completely insoluble.

i am thinking that i need to find the limiting reagent but we arnt given all of the products so i cant balance the equations

[nj_bartel]

What are the formulas for those two compounds?

[AWK]

Search your textbook, or wikipedia, or internet

[Borek]

Are you sure it is bromate and not bromide?

It is a classic double replacement reaction, other product should be obvious just from looking at anions and cations.

[n0m0r3l0v3_0k]

nAgNO3 = 0.79/170 = 79/17000
nKBrO3 = 0.473/167 = 473/167000

              AgNO3         +    KBrO3          -->      AgBrO3            +      KNO3
before     79/17000         473/167000                    0                          0
reaction  473/167000      473/167000             473/167000              473/167000
after      644/354875              0                    473/167000              473/167000

==>mAgBrO3 = (473/167000) x (108+80+16x3) = 0.6684311377 g
Is this right?

[Borek]

I don't like reagent selection, AgBrO₃ is not that weakly soluble, saturated solution has concentration of around 0.01M.

See output frame, bottom part of the window, below AgBrO₃.