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The moles of aluminum sulfate that could be produced from 32.7 grams of aluminum nitrate is 0.07 moles.
Relation between the mass and moles of any substance will be represented as:
n = W/M, where
W = given mass of aluminum nitrate = 32.7g
M = molar mass of aluminum nitrate = 213 g/mol
Moles of aluminum nitrate = 32.7 / 213 = 0.154 mol
Given chemical reaction is:
3Na₂SO₄(aq) + 2Al(NO₃)₃(aq) → Al₂(SO₄)₃(s) + 6NaNO₃(aq)
From the stoichiometry of the reaction it s clear that,
2 mole of Al(NO₃)₃ = produces 1 mole of Al₂(SO₄)₃
0.154 mole of Al(NO₃)₃ = produces (1/2)(0.154)=0.07 mole of Al₂(SO₄)₃
Hence produced moles of aluminum sulfate is 0.07 moles.
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