?$解：82.00\mathrm {g} \mathrm {MgSO}_{4}·7\mathrm {H}_{2}\mathrm {O}中含有硫酸鎂的質(zhì)量為 82.00 \mathrm {g}×\frac {24+32+16×4}{24+32+16×4+18×7}×100\%=40.00 \mathrm {g}$?
?$設(shè)菱鎂礦石中碳酸鎂的質(zhì)量為x。 $?
?$\mathrm {MgCO}_{3}+\mathrm {H}_{2}\mathrm {SO}_{4}\xlongequal[ ]{ }\mathrm {MgSO}_{4}+\mathrm {CO}_{2}↑+\mathrm {H}_{2}\mathrm {O} $?
?$\frac {84}{120}=\frac {x}{40.00\mathrm {g}} ，解得x=28.00\mathrm {g} $
?
?$菱鎂礦石的質(zhì)量為$?
?$\mathrm {m}(菱鎂礦石\mathrm {)}=\mathrm {m}(\mathrm {MgCO}_{3}\mathrm {)}+\mathrm {m}(\mathrm {MnCO}_{3}\mathrm {)}+\mathrm {m}(不溶物\mathrm {)}=28.00\mathrm {g}+2.30\mathrm {g}+4.70\mathrm {g}=35.00\mathrm {g} $
?
?$菱鎂礦石中碳酸鎂的質(zhì)量分?jǐn)?shù)為\frac {28.00\mathrm {g}}{35.00\mathrm {g}}×100\%=80\% $?
?$答：菱鎂礦石中碳酸鎂的質(zhì)量分?jǐn)?shù)為80\%。 $?
