diehell12
幼苗
共回答了14个问题采纳率:92.9% 举报
现计算出你配的溶液的摩尔浓度,n=m/M=1.5432g/106g/mol=0.0146mol,c=n/v=0.0146/0.25mol/l=0.0584mol/l,实际取出25ml,n=cv1=0.0584*0.025mol=0.00146mol
消耗的盐酸,n=c*v=0.1000*0.02468mol=0.002468mol,
盐酸和碳酸钠反应消耗比例是2:1,
那么参与反映的碳酸钠是0.002468/2mol=0.001234mol
含量,0.001234/0.00146=0.8452=85.2%
记得采纳
1年前
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