CSSA 2010 Chemistry Multiple Choice Question (1 Viewer)

[Dylanamali]

A lawn fertilser lists the sulfate content as 38.5% (w/w).

What mass of barium sulfate precipitate would be expected to form if a 1.50g sample of the fertiliser were analysed by reacting the sample with excess barium nitrate solution?

a) 0.238g
b) 0.578g
c) 1.40g
d) 3.64g

** I have no answer to this question but my answer was c). This is my working:

Ba(SO)4 -> Ba2+ + (SO4)2-
therefore Ba(SO)4 : (SO4)2- is 1:1
mass of sulfate = 1.50g x 0.385 = 0.5775g
n sulfate = 0.5775/96.07 = 0.006 mols
n sulfate = n barium sulfate = 0.006 mols
mass of barium sulfate = 0.006 x 233.37 = 1.40022 = 1.40g

Is this correct?

Thank you in advanced.

 

[Alkanes]

Everything looks right.

 

[s2 SEductive]

A lawn fertilser lists the sulfate content as 38.5% (w/w).

What mass of barium sulfate precipitate would be expected to form if a 1.50g sample of the fertiliser were analysed by reacting the sample with excess barium nitrate solution?

a) 0.238g
b) 0.578g
c) 1.40g
d) 3.64g

** I have no answer to this question but my answer was c). This is my working:

Ba(SO)4 -> Ba2+ + (SO4)2-
therefore Ba(SO)4 : (SO4)2- is 1:1
mass of sulfate = 1.50g x 0.385 = 0.5775g
n sulfate = 0.5775/96.07 = 0.006 mols
n sulfate = n barium sulfate = 0.006 mols
mass of barium sulfate = 0.006 x 233.37 = 1.40022 = 1.40g

Is this correct?

Thank you in advanced.

Yep correct
Btw here are the answers:
http://www.mediafire.com/?kwqw25niku7nk3c