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BeaMiranda
@BeaMiranda
January 2020
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Estequiometria.
3PbCl2 + Al2(SO4)3 --- 3PbSO4 + 2AlCl3
1) Quantas moléculas de AlCl3 são obtidas se usarmos 10 mols de PbCl2
2)Se usarmos 5 mols de Al2 (SO4)3 quantos gramas de PbSO4, obtemos?
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nandofilho10
3 PbCl2 + Al2(SO4)3 ---> 3PbSO4 + 2AlCl3
3 mol------------------------------------------------- 2mol
10 mol--------------------------------------------------x
x = 20 /3 mol
b) 3 PbCl2 + Al2(SO4)3 ---> 3PbSO4 + 2AlCl3
1 mol -----------------3mol
5mol------------------x
x = 15 mol
MM PbSO4 = 303 g / mol
15 mol x 303 g / mol =
4545 g
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BeaMiranda
Muito obrigado mesmo!
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3 mol------------------------------------------------- 2mol
10 mol--------------------------------------------------x
x = 20 /3 mol
b) 3 PbCl2 + Al2(SO4)3 ---> 3PbSO4 + 2AlCl3
1 mol -----------------3mol
5mol------------------x
x = 15 mol
MM PbSO4 = 303 g / mol
15 mol x 303 g / mol = 4545 g