Question

今有未知溶質(zhì)質(zhì)量分?jǐn)?shù)的鹽酸50 g，與7.8 g鋅充分反應(yīng)后，鋅有剩余，然后再加入20 g7.3%的鹽酸，恰好與剩余的鋅完全反應(yīng)，求：（1）第一次反應(yīng)時(shí)所用鹽酸中溶質(zhì)的質(zhì)量分?jǐn)?shù)．（2）反應(yīng)后所得溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)．

Answer 1

解：設(shè)與20 g7.3%的鹽酸反應(yīng)的鋅的質(zhì)量為X，反應(yīng)后生成的氫氣質(zhì)量為a₁，生成ZnCl₂質(zhì)量為b₁
Zn+2HCl═ZnCl₂+H₂↑
65 73
X 20 g×7.3%
=
X=1.3g
Zn+2HCl═ZnCl₂+H₂↑
73 136
20 g×7.3% b₁
=
b₁=2.72g
Zn+2HCl═ZnCl₂+H₂↑
73 2
20 g×7.3% a₁
=
a₁=0.04g

設(shè)第一次反應(yīng)時(shí)所用鹽酸中溶質(zhì)的質(zhì)量分?jǐn)?shù)為Y，反應(yīng)后生成的氫氣質(zhì)量為a₂，生成ZnCl₂質(zhì)量為b₂

Zn+2HCl═ZnCl₂+H₂↑
65 73 136 2
7.8 g-1.3g 50 gY b₂a₂
=
Y=14.6%
Zn+2HCl═ZnCl₂+H₂↑
65 136
7.8 g-1.3g b₂
=
b₂=13.6g
Zn+2HCl═ZnCl₂+H₂↑
65 2
7.8 g-1.3g a₂
=
a₂=0.2g
反應(yīng)后所得溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)為
×100%=21.0%
答：第一次反應(yīng)時(shí)所用鹽酸中溶質(zhì)的質(zhì)量分?jǐn)?shù)為14.6%；反應(yīng)后所得溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)為21.0%．
分析：計(jì)算第一次反應(yīng)時(shí)所用的鹽酸中溶質(zhì)的質(zhì)量分?jǐn)?shù)時(shí)，應(yīng)先計(jì)算出加入20 g7.3%的鹽酸時(shí)消耗的鋅的質(zhì)量，才能得出第一次所消耗的鋅的質(zhì)量，才能得出結(jié)果．反應(yīng)后所得溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)計(jì)算時(shí)，先計(jì)算出兩次反應(yīng)后所得溶質(zhì)的質(zhì)量，然后確定出所得溶液的質(zhì)量．
點(diǎn)評(píng)：在確定反應(yīng)后溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)時(shí)，一定要確定好反應(yīng)后的溶質(zhì)和溶液，否則就會(huì)得到錯(cuò)誤結(jié)果．