一个简单的化学题(In English)Which volume(体积)of sulphur dioxide(二氧化硫)(at r.t.p.) is formed when 9.7g of zinc sulphide(硫化锌) is heated in air?2ZnS + 3 O2(氧气) --> 2ZnO + 2 SO2(二氧化硫)A 1.2 dm3(立方分米) B 2.4 dm3 C 3.6dm3 D 4.8dm3

问题描述:

一个简单的化学题(In English)
Which volume(体积)of sulphur dioxide(二氧化硫)(at r.t.p.) is formed when 9.7g of zinc sulphide(硫化锌) is heated in air?
2ZnS + 3 O2(氧气) --> 2ZnO + 2 SO2(二氧化硫)
A 1.2 dm3(立方分米) B 2.4 dm3 C 3.6dm3 D 4.8dm3

计算过程:
分子量 ZnS 65+32=97
mol of ZnS = 9.7/97 = 0.1 mol
mol of SO2 = 0.1 mol
V of SO2 = 0.1 * 22.4 = 2.24 L = 2.24 dm3
比较近的是2.4 dm3
答案:B

The mole mass of zinc sulphide is 65+32=97g/molso the amount of zinc sulphide is 9.7/97=0.1mol2ZnS--2SO2 so the amount of sulphur dioxide is also 0.1molV=nVm=0.1mol*22.4L/mol=2.24L