Your answer of 3.75 would be correct if the limiting reagent were NaBr.
AgNO3+NaBr=AgBr↓+NaNO3
No. of moles of AgNO3 = 0.1*0.15 = 0.015 M
No. of moles of NaBr = 0.02*1 = 0.02 M
The limiting reagent is AgNO3, so the
No. of moles of AgBr produced = 0.015
Mass of AgBr
= 0.015 *(107.87+79.90)
= 2.82 g
what mass of solid AgBr is produced when 100.0 mL of 0.150 M AgNO3 is added to 20.0mL of 1.00 MNaBr?
i keep getting the answer 3.75 g however when i checked in the back of the book it gives me 2.82 g.. can you please explain the universal steps for me
2 answers
A 1.42 sample of a pure compound with formula M2SO4 was dissolved in water hand treated with an excess of aqueous calcium resulting in the precipitate of all the sulfate ions as calcium sulfate the precipitate was called dried and found to weight 1.36 determine the atomic mass of M and identify M.
2 answers