黑鞋白袜
幼苗
共回答了20个问题采纳率:85% 举报
1、设CO物质的量为x mol,CO2物质的量为y mol,则因为每mol气体的物质的量大约为22.4L,所以得
(x+y)·22.4=11.2
又二者质量合为20.4g,CO的分子量为28,CO2为44
28 x + 44 y =20.4
两个方程联立解得x =0.3 y = 0.2
所以混合气中二者质量比为:
0.3 ×28/0.2×44=21:22
设Na2CO3物质的量为n mol ,Na2SO4物质的量为m mol
两者分别与BaCl2反应的化学方程式如下:
Na2CO3 + BaCl2 =BaCO3 + 2NaCl
1 mol`````````````````1 mol
x mol`````````````````x mol
Na2SO4 + BaCl2 =BaSO4+ 2NaCl
1 mol``````````````````1 mol
y mol``````````````````y mol
可看出生成的沉淀为BaCO3 (分子量197)和BaSO4(分子量 233)
则 197·x + 233·y = 14.5
气体主要是BaCO3放出,方程式如下:
BaCO3 + 2HCl = BaCl2 + CO2 + H2O
1````````````````````````````````1
x````````````````````````````````x
则则x mol 的BaCO3 放出x mol的CO2,即22.4× x =1.12
x =0.05
将x =0.05 带入方程 197·x + 233·y ≈ 14.5 得 y =0.02
1年前
2