解：?$(1\mathrm {)}$?實(shí)驗(yàn)?$3$?和實(shí)驗(yàn)?$4$?中反應(yīng)的碳酸鈣有剩余。
?$(2\mathrm {)}$?由實(shí)驗(yàn)?$3$?可知，?$\mathrm {m}$?為?$4.4。$?
?$(3\mathrm {)}$?設(shè)這種石灰石中碳酸鈣的質(zhì)量分?jǐn)?shù)是?$x。$?
?$\mathrm {CaCO}_3+2\mathrm {HCl}\xlongequal[]{ }\mathrm {CaCl}_2+\mathrm {H}_2\mathrm {O}+\mathrm {CO}_2↑$?
100 44
10g×?$x$? 3.52g
?$\frac {100} {10\mathrm {g}×x }=\frac {44 }{3.52\mathrm {g} }，$?解得?$$ x = 80％
$$?
答：這種石灰石中碳酸鈣的質(zhì)量分?jǐn)?shù)為?$80\%。$?
