Question

某同學(xué)將40g的高錳酸鉀樣品（雜質(zhì)無(wú)氧氣放出）加熱制取氧氣，加熱到質(zhì)量不再變化后，停止加熱，冷卻稱(chēng)量剩余固體的質(zhì)量為36.8g．試求：                                        

（1）該同學(xué)制得氧氣的質(zhì)量是　．                                

（2）樣品中高錳酸鉀的質(zhì)量是多少？（寫(xiě)出計(jì)算過(guò)程）                  

Answer 1

【考點(diǎn)】根據(jù)化學(xué)反應(yīng)方程式的計(jì)算．                             

【專(zhuān)題】有關(guān)化學(xué)方程式的計(jì)算；有關(guān)化學(xué)方程式的計(jì)算．               

【分析】高錳酸鉀受熱分解生成錳酸鉀、二氧化錳和氧氣，反應(yīng)前后的質(zhì)量差即為反應(yīng)生成氧氣的質(zhì)量，根據(jù)氧氣的質(zhì)量可以計(jì)算高錳酸鉀的質(zhì)量．                                       

【解答】解：（1）該同學(xué)制得氧氣的質(zhì)量是：40g﹣36.8g=3.2g，         

故填：3.2g．                                                

（2）設(shè)樣品中高錳酸鉀的質(zhì)量為x，                               

2KMnO₄K₂MnO₄+MnO₂+O₂↑，                                  

316                  32                                        

x                   3.2g                                       

=，                                                

x=31.6g，                                                   

答：樣品中高錳酸鉀的質(zhì)量為31.6g．                              

【點(diǎn)評(píng)】差量法在計(jì)算中的應(yīng)用很廣泛，解答的關(guān)鍵是要分析出物質(zhì)的質(zhì)量差與要求的未知數(shù)之間的關(guān)系，再根據(jù)具體的數(shù)據(jù)求解．