高一化学问题，要过程

2Na2O2+2H2O=4NaOH+O2↑ 、Na2O+H2O=2NaOH
156 160 32 62 80
x y1 a 70-x y2
y1+y2=50％×(98+70-a)
160x/156+80(70-x)/62=0.5(168-32x/156)
x=39

（1）设减少的总质量，即生成的氧气的质量为x
{[(70 - x) *80/62] / (70 - x + 98) }*100% = 50%
x = 8 (g )
(2) 2Na2O2+2H2O=4NaOH+O2↑
156 g 32g
m(Na2O2) 8 g
m(Na2O2) = 39 g
(3 ) m(Na2O) = 70g - 39 g = 31 g

Na2O2 xmol Na2O ymol
78x+62y=70
80(x+y)=50%*(98+70-16x)