推荐回答（1个）
2A（气）←→xB（气）+C（气）
根据PV=nRT可得，当压强、物质的量一致时体积与温度成正比
673K，标准大气压的摩尔体积=22.4*673/273=55.22( L/mol)
混合气体密度为0.72g/L，所以平均分子量=0.72*55.22=39.76
混合气体总质量为46g,计算混合气体的物质的量=46g/)39.76g/mol=1.157mol
A占其体积百分比为58.84%,计算混合气体中A的物质的量=58.84%*1.157mol=0.68mol
（1）平衡混合气体的平均相对分子质量39.76
（2）x=2的值
混合气体的物质的量=1.157molA的物质的量=0.68mol,
根据A的变化量计算生成C=0.16mol
生成B=1.157mol-0.68mol-0.16mol=0.32mol
(恰好是C的2倍)所以B的系数是2
（3）A的平衡转化率0.68mol/1mol *100%=68%
标准状况通常指温度为0℃（273.15开）和压强为101.325千帕（1标准大气压，760毫米汞柱）的情况。使在比较气体体积时有统一的标准。气体的密度，除了特别说明的以外，都是指在标准状况下说的。
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