Question

15．6.5g鋅與100g稀硫酸恰好完全反應(yīng)，求所得溶液的溶質(zhì)的質(zhì)量分?jǐn)?shù)？

Answer 1

分析 欲求所得溶液的溶質(zhì)的質(zhì)量分?jǐn)?shù)，需先求生成的ZnSO₄和氫氣質(zhì)量，再根據(jù)溶質(zhì)質(zhì)量分?jǐn)?shù)=$\frac{溶質(zhì)質(zhì)量}{溶液質(zhì)量}$×100%計(jì)算即可．

解答 解：設(shè)生成氫氣和硫酸鋅的質(zhì)量分別為y、z
Zn+H₂SO₄=ZnSO₄+H₂↑
65        161  2
6.5g      z   y
$\frac{65}{6.5g}=\frac{2}{y}$，$\frac{65}{6.5g}=\frac{161}{z}$
解之得y=0.2g，z=16.1g
所得溶液的溶質(zhì)的質(zhì)量分?jǐn)?shù)
$\frac{16.1g}{6.5g+100g-0.2g}$×100%=15.1%
答：這種硫酸鋅的溶質(zhì)質(zhì)量分?jǐn)?shù)是15.1%

點(diǎn)評(píng) 本題主要考查學(xué)生對(duì)物質(zhì)質(zhì)量分?jǐn)?shù)和溶質(zhì)質(zhì)量分?jǐn)?shù)的計(jì)算能力．學(xué)生需正確書寫出化學(xué)方程式，并能正確運(yùn)用溶質(zhì)質(zhì)量分?jǐn)?shù)公式進(jìn)行計(jì)算，才能正確答題．