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Stoichiometry question?
One method of producing aluminum bromide is shown in the following unbalanced reaction. What mass of aluminum bromide can be produced by the reaction of sufficient aluminum sulfate with 8.75g of ammonium bromide?
Al2(SO4)3 + NH4Br --> AlBr3 + (NH4)2SO4
3 Answers
- Trevor HLv 79 years agoFavorite Answer
Write a properly balanced equation:
Al2(SO4)3 + 6 NH4Br = 2 AlBr3 + 3 (NH4)2SO4
6mol NH4Br will produce 2 mol AlBr3
Molar mass NH4Br = 97.94g/mol
8.75g NH4Br = 8.75/97.94 = 0.0893 mol NH4Br
From the balanced equation this will produce 0.0893*2/6 = 0.0298 mol AlBr3
Molar mass AlBr3 = 266.7g/mol
0.0298 mol = 0.0298*266.7 = 7.94 g AlBr3 produced.
- GianniLv 79 years ago
The balanced equation is:
Al2(SO4)3 + 6NH4Br --> 2AlBr3 + 3(NH4)2SO4
Now, you've got 8,75g of ammonium bromide. which corresponds to:
n(NH4Br) = 8,75g / 97,904g/mol = 0,0894mol
For each mole of NH4Br, you have a third of moles of AlBr3 ( ratio 6:3 ), then:
n(AlBr3) = n(NH4Br) = 0,0894mol * 1/3 = 0,0298mol ----->
------> m(AlBr3) = 0,0298mol * 266,7g/mol = 7,95g
Bye Bye :)
- harrie2harrieLv 79 years ago
Bal. Rxn.: Al2(SO4)3 + 6 NH4Br --> 2 AlBr3 + 3 (NH4)2SO4
2 moles of AlBr3 are formed per 6 moles of NH4Br (1:2)
moles NH4Br: 8.75 g/molec wt NH4Br
? moles NH4Br x 1/2 x molec. wt AlBr3= answer
