Question

If 49.0mL of 0.250 M AgC2H3O2 are mixed with 53.5 mL of 0.600 M Na2CrO4, how many of Ag2CrO4 (331.7 g/mol) are produced,assuming 100% yield?
given this reaction: 2AgC2H3O2+Na2CrO4---> AgCrO4 + 2NaC2H3O2

a)10.65g
b)7.19g
c)2.74g
d)2.03g

Answer 1

1) With the volume in L and the concentration you can calculate the number of moles of the two reactants. n= M x V(L)
Convert the volume from mL to L and multiply by the molarity.
the equation has a typo mistake in the AgCrO4 should be Ag2CrO4
2AgC2H3O2+Na2CrO4---> Ag2CrO4 + 2NaC2H3O2

2) The stoiquiometry of the reaction is telling that for
2 moles of AgC2H3O2 you get one mole of Ag2CrO4,
and for 1 mole of Na2CrO4 you get one mole of Ag2CrO4

calculate the limiting reactant by dividing the moles of AgC2H3O2 by 2 and compare this value with the moles of Na2CrO4, the smallest number is the limiting reactant and will be the number of moles of Ag2CrO4 that you can obtain.

3) multiply the number of moles of the Ag2CrO4 by its molecular mass (331.7 g/mol) and you will get the g of Ag2CrO4