Finish the following lab, will reward more point or partial point on how much yo

Question

Finish the following lab, will reward more point or partial point on how much you do, if you do a good job ect.

Precipitation of Ba3(PO4)2 from the Salt Mixture Unknown number ________ Trial 1 Trial 2 Tared mass of salt mixture (g) Mass of filter paper (g) Mass of filter paper and Ba3(PO4)2 (g) Mass of Ba3(PO4)2 (g) Determination of Limiting Reactant Limiting reactant in salt mixture (write complete formula) Excess reactant in salt mixture (write complete formula) Data Analysis Moles of Ba3(PO4)2 precipitated (mol) Moles of limiting reactant in salt mixture (mol) formula of limiting hydrate Mass of limiting reactant in salt mixture (g) formula of limiting hydrate Mass of excess reactant in salt mixture (g) formula of excess hydrate Percent limiting reactant in salt mixture (%) formula of limiting hydrate Percent excess reactant in salt mixture (%) formula of excess hydrate

Explanation / Answer

i m using mean of both trial so tare mass of salt mixture = .997+.998 /2 =.9975 g

mass of filter paper =1.253+ 1.243 / 2= 1.248g

mass of filter paper and Ba3(po4)2 = 1.64 + 1.52 /2 = 1.68gm

mass of Ba3(po4)2 = 1.68 - 1.24 = .44 gm

moles of Ba3(po4)2 :-

Barium has atomic mass of 137.38.

phosphorus has a mass of 30.97.
and Oxygen has a mass of 16.

Since there are 3 Barium atoms:
3x137.38 = 412.14

Distribute the 2 at the end of (PO4)2

So Phosphorus has 2 atoms now.
30.97x2 = 61.94

After distributing Oxygen has 8.
So 8x16 = 128.

Add the three up:
412.14 +61.94 + 128 = 602.08 ,

percentage of phosphorus(limiting agent) in mass of salt mixture = 189.94 / 602.08 = 31.94 %

percentage of barium(excess agent) in mass of salt mixture = 412.14 / 602.08 = 68.45%

The real Molar mass of Barium Phospate is 602.08 g

mole of Barium Phospate :-

mass of Barium Phospate given / tare mass of salt mixture = .9975 / 602.08= .9975 * 10 ^-3 mol

moles of limiting agent:-

limiting agent = po4^-2

mass of limiting agent in given mass of Barium Phospate = .9975 * 31.94 / 100 = .318 g

moles of limiting agent = .318 / 189.94 =1.67 * 10 ^-3 mol

mass of excess agent in given mass of Barium Phospate= .9975 * 68.45 /100 = .682 g

moles of exess agent = .682 / 412.14 = 1.65 * 10^-3 mol

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