lervei
幼苗
共回答了12个问题采纳率:100% 举报
D
8.34g FeS0
4 •7H
2 0样品物质的量n=8.34g/278g/mol=0.03mol,其中m(H
2 0)=0.03mol×7×18g/mol=3.78g,如晶体全部失去结晶水,固体的质量应为8.34g-3.78g=4.56g,可知在加热到373℃之前,晶体失去部分结晶水。、
A.温度为78℃时,固体质量为6.72g,其中m(FeS0
4 )=0.03mol×152g/mol=4.56g,m(H20)=6.72g-4.56g=2.16g,n(H20)=2.16g/18g/mol==0.12mol,则n(H
2 0):n(FeS0
4 )=0.12mol:0.03mol=4:1,则化学式为FeSO
4 •
4 H
2 O,故A错误;B.温度为l59℃时,固体质量为5.10g,其中m(FeS0
4 )=0.03mol×152g/mol=4.56g,m(H
2 0)=5.10g-4.56g=0.54g,n(H
2 0)=0.54g/18g/mol=0.03mol,则n(H
2 0):n(FeS0
4 )=0.03mol:0.03mol=1:1,则化学式为FeSO
4 •H2O,故B错误;
C.m(FeS0
4 )=0.03mol×152g/mol=4.56g,则在隔绝空气条件下由N得到P的化学方程式为FeSO
4 •H
2 O
![](https://img.yulucn.com/upload/c/fd/cfd67564cd5561cb255e261fa9b13db8_thumb.jpg)
FeS0
4 +H
2 0,故C错误;
D.加热至633℃时,固体的质量为2.40g,其中n(Fe)=n( FeS0
4 •7H
2 0)=0.03mol,m(Fe)=0.03mol×56g/mol=1.68g,则固体中m(O)=2.40g-1.68g=0.72g,n(O)=0.72g/16g/mo=0.045mol,则n(Fe):n(O)=0.03mol:0.045mol=2:3,则固体物质Q的化学式为Fe
2 O
3 ,故D正确
1年前
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