Помогите решить, безотлагательно необходимо! Задачка. Технический алюминий массой 33,75 г с

дано

m техн(Al) = 33.75 g

W(прим) = 20%

m(ppaH2SO4) = 784 g

W(H2SO4) = 15%

--------------------------

m(Al2(SO4)3)-?

m чист(Al) = 33.75 - (33.75*20% / 100%) = 27 g

M(Al) = 27 g/mol

n(Al) = m(Al) / M(AL) = 27  /27 = 1 mol

m(H2SO4) = 784 * 15% / 100% = 117.6 g

M(H2SO4) = 98 g/mol

n(H2SO4) = m(HsSO4) / M(H2SO4) = 117.6 / 98 = 1.2 mol

n(Al) lt; n(H2SO4

2Al+3H2SO4--gt;Al2(SO4)3+3H2

M(Al2(SO4)3) = 278 g/mol

2n(Al) = n(Al2(SO4)3) = 1 mol

n(AL2(SO4)3) = 1/2 = 0.5 mol

m(Al2(SO4)3) = n*M = 0.5*278 = 139 g

ответ 139 г