初2化学题(急啊,明天我们还要用,请快答啊)
来源:百度知道 编辑:UC知道 时间:2024/05/16 02:53:53
这个题目的原题是要求出具体的数
要使天平平衡,左右实际增加的质量要相等.
Fe+H2SO4=FeSO4+H2↑实际增重
56``````````````````2``````54
5.6`````````````````0.2`````5.4
Zn+H2SO4=ZnSO4+H2↑实际增重
65``````````````````2``````63
x````````````````````````````5.4
x=5.57g
Fe+H2SO4=FeSO4+H2↑
56 2
5.6 0.2
Zn+H2SO4=ZnSO4+H2↑
65 2
X 0.2
解得X=6.5
Fe+H2SO4=FeSO4+H2↑
56 2
5.6 0.2
Zn+H2SO4=ZnSO4+H2↑
65 2
X 0.2
所以X=6.5,因为要右偏,所以大于6.5g
解:设加入的金属锌的质量是m
根据反应后溶液的质量相等计算
Fe+H2SO4=FeSO4+H2↑
56 2
5.6 mH2
56/5.6=2/mH2
∴mH2=0.2g
Zn+H2SO4=ZnSO4+H2↑
65 2
m mH2′
65/m=2/mH2′
∴mH2′=2m/65
mFe+mH2SO4=m液+mH2
mZn+mH2SO4=m液+mH2′
∴mFe-mH2=mZn-mH2′
5.6-0.2 g=m-(2m/65)
5.4=63m/65
∴m≈5.56g
答:右边应加锌的质量为5.56g
Fe+H2SO4=FeSO4+H2↑
56 2
5.6 0.2 实际增重为5.6-0.2=5