Question

為測(cè)定某石灰石樣品中碳酸鈣的含量，小麗進(jìn)行實(shí)驗(yàn)，測(cè)得相關(guān)數(shù)據(jù)如下圖所示：           

           

（1）完全反應(yīng)生成二氧化碳的質(zhì)量為　　．                        

（2）該石灰石樣品中碳酸鈣的質(zhì)量分?jǐn)?shù)是多少？                        

（3）反應(yīng)后溶質(zhì)質(zhì)量分?jǐn)?shù)是多少？                                

Answer 1

【考點(diǎn)】根據(jù)化學(xué)反應(yīng)方程式的計(jì)算；有關(guān)溶質(zhì)質(zhì)量分?jǐn)?shù)的簡(jiǎn)單計(jì)算．         

【專題】溶質(zhì)質(zhì)量分?jǐn)?shù)與化學(xué)方程式相結(jié)合的計(jì)算．                     

【分析】根據(jù)碳酸鈣和稀鹽酸反應(yīng)生成氯化鈣、水和二氧化碳，反應(yīng)前后的質(zhì)量差即為反應(yīng)生成二氧化碳質(zhì)量，根據(jù)二氧化碳質(zhì)量可以計(jì)算碳酸鈣質(zhì)量和生成氯化鈣的質(zhì)量，進(jìn)一步可以計(jì)算該石灰石樣品中碳酸鈣的質(zhì)量分?jǐn)?shù)和反應(yīng)后溶質(zhì)質(zhì)量分?jǐn)?shù)．                                                      

【解答】解：（1）完全反應(yīng)生成二氧化碳的質(zhì)量為：40g+200g﹣231.2g=8.8g；         

（2）設(shè)40g石灰石樣品中碳酸鈣的質(zhì)量為x，生成氯化鈣的質(zhì)量為y，         

CaCO₃+2HCl=CaCl₂+H₂O+CO₂↑                                       

100        111       44                                        

x           y        8.8g                                      

==                                              

x=20g                                                       

y=22.2g                                                     

石灰石樣品中碳酸鈣的質(zhì)量分?jǐn)?shù)為：×100%=50%；                   

（3）反應(yīng)后溶質(zhì)質(zhì)量分?jǐn)?shù)是：×100%=10.5%            

故答案為：（1）8.8g；                                          

（2）石灰石樣品中碳酸鈣的質(zhì)量分?jǐn)?shù)為50%；                           

（3）反應(yīng)后溶質(zhì)質(zhì)量分?jǐn)?shù)是10.5%．                               

【點(diǎn)評(píng)】本題主要考查學(xué)生運(yùn)用假設(shè)法和化學(xué)方程式進(jìn)行計(jì)算和推斷的能力，計(jì)算時(shí)要注意規(guī)范性和準(zhǔn)確性．