дано
m(NaOH) = 2.8 g
m(Al2(SO4)3) = 3.42 g
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m(AL(OH)3)-?
M(NaOH) = 40 g/mol
n(NaOH) = m/M = 2.8 / 40 = 0.07 mol
M(Al2(SO4)3) = 342 g/mol
n(Al2(SO4)3) = m/M =3.42 / 342 = 0.01 mol
n(NaOH) > n(AL2(SO4)3)
6NaOH+Al2(SO4)3-->2Al(OH)3+3Na2SO4
n(AL2(SO4)3) = 2n(Al(OH)3) = 0.01 mol
M(Al(OH)3) = 78 g/mol
m(Al(OH)3) = n*M = 0.01 * 78 = 0.78 g
ответ 0.78 г
