stoichiometry! ! !?

Question

a double replacement reaction occurs between silver nitrate and magnesium bromide.calculate the mass of silver bromide which should be produced from 22.5g of silver nitrate.
When reacted in the lab, 22.0 grams of silver bromide were actually produce. What is the percent yield?

-how do i do this?? and whats the answer? =/ ((i think i did it wrong i got 88%))

Answer 1

All you have to do for percent yield is..

22.0g (actual yeild)
----------------------------- x100
25.0g (theoretical yield)

actual/theoretical x100

Answer 2

Looks right to me.
AgNO3 +½MgBr2 -> AgBr + ½Mg(NO3)2
or equally acceptable
2AgNO3 + MgBr2 -> 2AgBr + Mg(No3)2
so figure the formula weight of each silver species
22.5g of silver nitrate is 22.5 ÷ F.W.sn = moles sn
22g of AgBr is 22 ÷ F.W.sb = moles of sb (or if you're particular "equivalents", not moles, of the salts)
but according to the stoichiometry its 1:1 or 2:2
so every mole going in, if recovered should go out to give you 100% yield. IF this happens you know someone cheated or was sloppy since the real world is not perfect and you never get 100% yield. Out ÷ In (moles) gives predicted yield or more accurately Out ÷ Out(100%) gives yield. Be sure to convert fraction eg .8846 to percent eg 88.5%. Looks right to me.

Answer 3

2 AgNO3 + MgBr2 ---> 2 AgBr + Mg(NO3)2

Assuming you have excess MgBr2, AgNO3 is your limiting reactant.

Theoretical yield = 22.5 g AgNO3 * 1 mole/169.86 g/mole * 2 moles AgBr/2 moles AgNO3 * 186.76 g AgBr/mole = 24.74 g AgBr

% yield = actual/theoretical * 100 = 22.0/24.74 * 100 = 88.9%