化学高手进！

1.n(H2):n(o2)=2:1

M=2*2/3+32*1/3=12

p=M/V

2.W=36.5PV/(36.5V+18);C=1000PW/36.5把W带入即可

3.假设合金全为Mg，则Mg-------H2

0.125mol
m(Mg）=24*0.125=3g

假设全为Fe，则
Fe-----H2

m（Fe）=0.125*56=7g

合金的质量3g

1.

n(H2)=2g/2g/mol=1mol

n(O2)=16g/32g/mol=0,5mol

M
=(2g+16g)/(1mol+0.5mol)=12g/mol

只可以算出标准状况下的密度
p=12g/mol/22.4L/mol=0.536g/L

2.假设是1L水中溶解VLHCl

n(HCl)=V/22.4
mol

V(aq)=(36.5V/22.4+1000)/(1000p)

c(HCl)=V/22.4
/[(36.5V/22.4+1000)/(1000p)]

=1000pV/(36.5V+22400)
mol/L

w(HCl)=36.5V/22.4
/(36.5V/22.4+1000)*100%

=36.5V/(36.5V
+22400)*100%