解：設(shè)純堿樣品中碳酸鈉的質(zhì)量為?$x，$?稀鹽酸中氯化氫的質(zhì)量為?$y，$?反應(yīng)生成的
氯化鈉的質(zhì)量為?$z。$?
?$\ \text {Na}_2\ \text {CO}_3+2\ \text {HCl}\xlongequal[]{}2\ \text {NaCl}+\ \text H_2\ \text {O}+\ \text {CO}_2↑$?
106 73 117 44
?$x$? y z 4.4 g
?$\mathrm{ \frac { 106 }{ x }=\frac { 73 }{y }=\frac { 117 }{ z }=\frac { 44 }{ 4.4g } }，$??$ $?解得?$ x = 10.6 \mathrm {g}，$??$y = 7.3 \mathrm {g}，$??$z = 11.7 \mathrm {g}$?
?$(1\mathrm {)}$?純堿中?$ \mathrm{ Na_2CO_3} $?的質(zhì)量分數(shù)為?$ \mathrm{\frac{ 10.6g}{ 11.4g}×100％= 93％}$?
?$(2\mathrm {)}$?反應(yīng)前稀鹽酸的質(zhì)量為?$7.3\mathrm {g} ÷ 7.3％ = 100\mathrm {g}$?
反應(yīng)后溶液中溶質(zhì)的質(zhì)量分數(shù)為
?$\mathrm{\frac{11.7 g+(11.4 g-10.6 g)}{100 g+104.4 g-4.4 g} ×100 \%=6.25 \% }$?
答：
?$(1\mathrm {)}$?純堿中碳酸鈉的質(zhì)量分數(shù)是?$93.0\%；$?
?$(2\mathrm {)}$?反應(yīng)后溶液中溶質(zhì)的質(zhì)量分數(shù)是?$6.25\%。$?
