Question

小女子在这里谢过了将m 克铜屑、铁屑的混合物溶解在过量稀硝酸中,得到NO气体（标准状况下）的体积v L；向上述溶液中加入

Answer 1

3Cu(s)+8HNO3(a.q.)=3Cu(NO3)2(a.q.)+2NO(g)+4H2O

化合价升降配平法Cu失2个电子,N得3个电子,失电子数等于得电子数等于2和3的最小公倍数6

Cu前配3,Cu(NO3)2前配3,参与反应的NO3-配2,加上Cu(NO3)的NO3-所以HNO3前面配8

最后水配4

Fe(s)+4HNO3(a.q.)=Fe(NO3)3(a.q.)+NO(g)+2H2O

配平同上

中英文切着烦,全部用英文写了

Suppose Fe is x mol,Cu is y mol

56x+64y=m (1)

Fe(s)+4HNO3(a.q.)=Fe(NO3)3(a.q.)+NO(g)+2H2O

x mol x mol

3Cu(s)+8HNO3(a.q.)=3Cu(NO3)2(a.q.)+2NO(g)+4H2O

y mol 2·y/3 mol

Under the standard 1 atm pressure 1 mol gas makes 22.4L volume

So

22.4·x+22.4·(2·y/3)=v (2)

According to (1)&(2)

{

x=(4.29V-m)/40

y=(m-2.5V)/26.7

}

The mass of the dross is

x·107+y·98 g

Warning:No calculation insurance,you'd better try it again!

By using the amount of the substance it is easier to switch between the mass and the volume.