cpa_sunjun
花朵
共回答了17个问题采纳率:94.1% 举报
答:
1.混合后 w = (35*0.1 + 65*0.20/(35+65)*100% = 16.5%
2.HCl ------ NaOH
36.5g------40g
16.5%*50---20%*m(NaOH)
得 m(NaOH) = 16.5% *50*40 /36.5/20% = 45.21g
3.反应后溶液中溶质为NaCl,其物质的量与HCl的量相同,溶液总质量守恒,故为(50+45.21)=95.21g
则 m(NaCl) = 16.5%*50/36.5*58.5 = 13.22g
w(NaCl) = 13.22/95.21 *100% = 13.9%
综上所述,1混合后盐酸中溶质的质量分数为16.5%;2.所用NaOH溶液的质量 为45.21g;
3.反应后溶液中溶质的质量分数 13.9%
1年前
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