Question

初三化学计算题！急急急求计算过程！高手来

为测定某一黄铜样品中锌的质量分数，某同学称取20g黄铜（铜和锌的合金）样品于烧杯中，向其中加入稀硫酸至不在产生气泡为止，共消耗100g质量分数为9.8百分之，计算：（1）该黄铜中锌的质量。（2）反应后所得溶液的容质质量分数。（计算结果保留0，1百分之）
自己在家里学习，找谁问呢，我卷子上题目就这样吧，好像没漏字

Answer 1

设锌的质量为x，生成的硫酸锌的质量为y，H2质量为m，根据题意有：H2SO4的质量为100g*9.8%=9.8g
Zn+H2SO4==ZnSO4+H2
65 98 161 2
x 9.8g y m
然后列比列式并求解可得：x=6.5g
y=16.1g
m=0.2g
（1）合金中锌的质量为6.5g；
（2）所得硫酸锌溶液中溶质的质量分数为：16.1g/（100g+6.5g-0.2g）=15.1%

Answer 2

Subject did not say clearly , Dropouts , though I learned chemistry , learn
well, and the University is chemistry , but do subject or tired , you write it
well , I'll give you to do it, in fact, this topic and the teacher asked the
students best , easy , why do online , say , who are willing to give you title
it exactly right downstairs , 15.1%

Answer 3

(1)m(Zn)=6.5
(2)16.1/(100-9.8+16.1)=15.1%

Answer 4

想想