将13克不纯的锌粒投入盛有100克稀硫酸的烧杯中,恰好完全反应,生成氢气0.4克计算
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Zn-2H2,0.4gH2=0.2mol

故锌0.1mol,纯度0.1*65/13=50%

0.2mol硫酸,0.2*98/100=19.6%