First determine the actual yield, moles of Sr 3 (PO 4 ) 2 formed. From the actua

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!

Explanation / Answer

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 %

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