$解:设所加盐酸的溶质质量分数为\mathrm {x}_{}。$
$\mathrm {NaHCO}_3+\mathrm {HCl}\xlongequal[]{}\mathrm {NaCl}+\mathrm {H}_2\mathrm {O}+\mathrm {CO}_2↑$
36.5 44
100g×x 4.4g
$\mathrm{ \frac {36.5 }{100g×x }=\frac {44 }{4.4g } } $ $解得 \mathrm {x}_{} = 3.65%$
$答:所加盐酸的溶质质量分数为3.65%。$