(1)n(O 2)=
0.0672ml
22.4L/mol =0.003mol,
2(CaO 2?xH 2O)
△
.
2CaO+O 2↑+2xH 2O
2mol 1mol
n(CaO 2?xH 2O)0.003mol
n(CaO 2?xH 2O)=0.006mol,
则n(CaO 2)=n(CaO 2?xH 2O)=0.006mol,
故答案为:0.006mol;
(2)n(CaCO 3)=
0.70g
100g/mol =0.007mol,
①根据Ca元素守恒,可知:n(CaO)=0.007mol-0.006mol=0.001mol,
m(CaO)=0.001mol×56g/mol=0.056g,
答:样品中CaO的质量为0.056g;
②样品中水的质量为:m(H 2O)=0.542g-m(CaO 2)-m(CaO)=0.542g-0.006mol×72g/mol-0.056g=0.054g,
n(H 2O)=
0.054g
18g/mol =0.003mol,
x=
n (H 2 O)
n(C aO 2 ) =
0.003mol
0.006mol =0.5,
答:样品中CaO 2?xH 2O的x值为0.5.