A 0.649 g sample containing only K2SO4 (M.M. 174.25) and (NH4)2SO4 (M.M. 132.13) was dissolved in water and treated with Ba(NO3)2 to participate all SO42- as BaSO4 (M.M. 233.38). Find the weight percent of K2SO4 in the sample if 0.977 g of precipitate was formed.

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mickymike92 mickymike92 · Answer 1 · 2022-03-16T18:25:34+01:00

Based in the data provided, the percentage by mass of K2SO4 is 60.55 %.

The mass of K2SO4 in the mixture is obtained from the data given as follows:

Total mass of sample = 0.649 g

Let the mass of K2SO4 in the mixture be X g

Mass (NH4)SO4 = (0.649 - X )g

Moles of K2SO4 = X g/174.25 g/mol

= 0.00574X mol

Moles of (NH4)2SO4 = (0.649 - X)g / 132.13 g/mol

= (0.00491 - 0.00757X) mol

Total number of moles SO42- ions precipitated = 0.00574X mol + (0.00491 - 0.00757X)mol

= (0.00491-0.00183X) mol

Mole ratio of SO42- ions to Ba2+ ions = 1:1

Thus, moles of Ba2+ ions = (0.00491 - 0.00183X)mol

Moles of BaSO4 precipitated = 0.977 g/233.38 g/mol = 0.00419 moles

1 mole of BaSO4 produces 1 mole of Ba2+ ions

Therefore,

(0.00491 - 0.00183X) mol = 0.00419mol

0.00183X = 0.00491 - 0.00419mol

0.00183X = 0.00072

X = 0.00072/0.00183 = 0.393 g

Mass of K2SO4 = 0.393 g

Percentage by mass of K2SO4 = (0.393/0.649) × 100

Percentage by mass of K2SO4 = 60.55 %

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