Question

Помогите пожалуйста!!!

Answer 1

дано

m(Al(OH)3) =300 g

w(Na3AlO3) = 95%

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

m(Na3ALO3)-?

2Al(OH)3+3Na2O-->2Na3AlO3+3H2O

M(Al(OH)3)= 78 g/mol

n(Al(OH)3) = m/M = 300 / 78 = 3.85 mol

2n(Al(OH)3) = 2n(Na3AlO3) = 3.85mol

M(Na3AlO3) = 144 g/mol

m теор (Na3AlO3) = n*M = 3.85 * 144 = 554.4 g

m пр(Na3AlO3) = mтеор(Na3AlO3) * W(Na3AlO3) / 100% = 554.4*95% / 100%

m пр(Na3AlO3) = 526.7 g

ответ 526.7 g