Chemistry help! ASAP?

Pinky! · #1 (**permalink**) 03-14-2008, 08:42 PM

What mass of barium chloride will be needed to react completely with 113g of aluminum sulfate to produce barium sulfate and aluminum chloride?
Please show me the steps

Fadia · #2 (**permalink**) 03-14-2008, 08:42 PM

The balanced equation of the reaction is:
3BaCL2 + AL2(SO4)3 -----> 3BaSO4 +2 ALCl3
moles of AL2(SO4)3 = 113 g / 342 (g/mol)=0.33 mol
moles of 3BaCL2 = 0.33*3 =0.99
mass of 3BaCL2 = 0.99*208.22= 206 g

Ohmshri · #3 (**permalink**) 03-14-2008, 08:43 PM

Molecular weights
Al2(SO4) = 342.15 g/mol for anhydridous salt
AlCl3 = 133.34 g/mol for anhydridous salt
BaCl2 = 208.2324 g/mol for anhydridous salt
BaSO4 = 233.43 g/mol for anhydridous salt

reaction
3*n*BaCl2+n*Al2(SO4)3 = 2*n*AlCl3+3*n*Ba(SO4)

n=113/342.15
m=3*n*(BaCl2)
m=3*(113/342.15)*208.2324
m=206.32g

husnitahahusnitaha · #4 (**permalink**) 03-14-2008, 08:43 PM

Al2(SO4)3 + 2BaCl3-------2AlCl3 + Ba2(SO4)3
1Mole need 2Mols
340grams need 235 grams
113 need x
x= 235x113/340= 78.1 grams of Barium chloride needed

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