Chem question (1 Viewer)

[Bobbo1]

In order to determine the sulfate content of a commercial lawn-food, a student
dissolves a 4.58 g sample of the lawn-food, adds excess BaCl2(aq), filters the
mixture, and finally washes and dries the residue.
After drying, the residue had a mass of 7.95g.
What is the % sulfate in the lawnfood, and how could the student have made their
experiment more reliable?
% sulfate To increase reliability:
(A) 58 Repeat the experiment and average the results.
(B) 73 Repeat the experiment and average the results.
(C) 58 Control important variables during the experiment.
(D) 73 Control important variables during the experiment.

I keep getting 23.85%, so can someone help me out?

 

[brettymaccc]

The 4.58g sample contains the sulfate, and upon adding BaCl2, BaSO4 is precipitated out (which has the mass of 7.95g).

Ba + SO4 = BaSO4.

Number of moles of BaSO4 reacted = 7.95g * (1 mol/233.37g), which equals 3.4*10^-2 mol. [I just used the gram formula weight of Barium Sulfate to get the number of moles].

Thus, from the equation, x mol of BaSO4 is produced from x mol of SO4. Thus, 3.4*10^-2 mol of BaSO4 is produced from 3.4*10^-2 mol of SO4.

Mass of sulfate in sample = 3.4*10^-2 mol * (96.07g/1mol) = 3.266g.

Thus, percentage by mass = 3.266/4.58 * 100 = 71.3%, I'd say B to increase reliability and because our answer is closer to 73% than to 58%.

 

[Bobbo1]

The 4.58g sample contains the sulfate, and upon adding BaCl2, BaSO4 is precipitated out (which has the mass of 7.95g).

Ba + SO4 = BaSO4.

Number of moles of BaSO4 reacted = 7.95g * (1 mol/233.37g), which equals 3.4*10^-2 mol. [I just used the gram formula weight of Barium Sulfate to get the number of moles].

Thus, from the equation, x mol of BaSO4 is produced from x mol of SO4. Thus, 3.4*10^-2 mol of BaSO4 is produced from 3.4*10^-2 mol of SO4.

Mass of sulfate in sample = 3.4*10^-2 mol * (96.07g/1mol) = 3.266g.

Thus, percentage by mass = 3.266/4.58 * 100 = 71.3%, I'd say B to increase reliability and because our answer is closer to 73% than to 58%.

ty man, had a complete mind blank and did something stupid for this question - probably need to revise it myself