дано
W(O2) = 61.5%
Me(OH)3
----------------------
Me-?
W(O2) = Ar(O)*n/ M(Me(OH)3) * 100%
61.5% = 16*3 / (X*+51) * 100%
61.5X + 3136.5=4800
X = 27 -- Al
M(AL(OH)3) = 27 + (16+1)*3 = 78 g/mol
ответ АЛЮМИНИЙ, 78 г/моль
2)
Дано
n(NaOH) = 0.3 mol
n(Ca(OH)2) = 0.4 mol
-----------------------------
m(см)-?
M(NaOH) = 40 g/mol
m(NaOH)= n*M = 0.3*40 = 12 g
M(Ca(OH)2 )= 74 g/mol
m(Ca(OH)2) = n*M = 0.4*74=29.6 g
m(см) = m(NaOH)+m(Ca(OH)2) = 12+29.6 = 41.6 g
ответ 41.6 г