дано
m(Al(OH)3) =300 g
w(Na3AlO3) = 95%
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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