粗铵盐1.000g加过量NaOH溶液，产生的氨经蒸汽吸收在50.00ml,0.5000mol/L的酸中

显示全部楼层 · 2012-9-25 14:33:24

过量酸的物质的量=NaOH的物质的量=0.5*1.56*0.001=0.00078mol氨的物质的量=50*0.001*0.5-0.00078=0.02422mol粗铵盐中NH3的百分含量=(0.02422*17)/1*100%=41.174%...

千问 · 2012-9-25 14:33:24

过量酸的物质的量=NaOH的物质的量=0.5*1.56*0.001=0.00078mol氨的物质的量=50*0.001*0.5-0.00078=0.02422mol粗铵盐中NH3的百分含量=(0.02422*17)/1*100%=41.174% 两者之间一比一...

千问 · 2012-9-25 14:33:24

过量的酸n1(H+)=0.5X1.56/1000=0.78/1000 mol与氨反应的酸n2(H+)=(50X0.5-0.78)/1000=24.22/1000moln(NH3)=24.22/1000 molW=n(NH3)XM(NH3)X100%/1=41.2%...