Question

First determine the actual yield, moles of Sr₃(PO₄)₂ formed. From the actual yield, theoretical moles of limiting reactant in sample:

a) Theoretical mass of limiting reactant

b) Mass percent of limiting reactant in sample

From the actual yield, mass of excess reactant sample, what is the mass percent of excess in sample?

Beaker Sample II

Unknown mixture NaPO412H2O * SrCl2* 6H2O

Mass of mixture and container 113.4904g

Mass of container 111.5575g

Mass of mixture added to Beaker II 1.9329g

Formula of limiting reactant Na3PO4

Formula of excess reactant SrCl2

Mass of paper and precipitate: First weighing 0.9486g, second weighing 0.9486g

Mass of filter paper .6774g

Thank you!

Answer 1

answer:

mass of mixture + container = 113.4904 gm

mass of container = 111.5575 gm

mass of mixture = 113.4904 gm -111.5575 gm =  1.9329 gm

now the standard reaction will be

2Na₃PO₄ + 3SrCl₂ = Sr₃(PO₄)₂ + 6NaCl [balanced equation]

limiting precipitate

(product)

Mass of paper + precipitate = 0.9486 gm

Mass of paper = 0.6774 gm

Mass of precipitate = 0.9468 gm - 0.6774 gm =0.2694 gm

hence, actual yield of the product will be 0.2694 gm

moles of Sr₃(PO₄)₂ = (0.2694 gm /452.8027 gm) = 5.94*10^-4   [ molecular weight of Sr₃(PO₄)₂ =452.8027 gm]

from stoichiometric coefficient values of balanced equation it is clear that

1 mol Sr₃(PO₄)₂ is produced from 2 mol of limiting reagent Na₃PO₄

hence,5.94*10^-4 mol Sr₃(PO₄)₂ will produced from = 2*5.94*10^-4 = 1.18*10^-3 mol Na₃PO₄

a) theoretical mass of limiting reactant is = moles * molecular weight = 1.18*10^-3 mol * 163.94 gm/mol = 0.1934 gm

b) mass percent of limiting reagent(%) = (0.1934 gm/1.9329 gm) * 100 = 10%

from stoichiometric coefficient values of balanced equation it is clear that

2 mol Na₃PO₄ is reacted with 3 mol of SrCl₂

1.18*10^-3 mol Na₃PO₄ will reacted with = 1.77*10^-3 mol of SrCl₂

so, weight of 1.77*10^-3 mol of SrCl₂ = 1.77*10^-3 mol * 158.53 gm/mol =0.2805 gm

mass of the reacted sample = weight of Na₃PO₄ + weight of reacted SrCl₂ =0.1934 gm + 0.2805 gm = 0.4739 gm

mass of excess reactant sample = mass of the mixture - mass of the reacted sample = 1.9329 gm - 0.4739 gm = 1.459 gm

mass percent of excess reactant sample =[ (1.459 gm /1.9329 gm)*100] % = 75.48 %