標題:
chemistry 問題
發問:
0.92g of a mixture of metallic copper and copper (II) oxide was heated in a stream of hydrogen until reduction was copplete. the residure weighed 0.752 g. what was the percentage of metallic copper in the original mixture? ANS::::Molar mass of Cu = 63.5 g mol-1 (in some books, 64 g mol-1)Molar mass of CuO... 顯示更多 0.92g of a mixture of metallic copper and copper (II) oxide was heated in a stream of hydrogen until reduction was copplete. the residure weighed 0.752 g. what was the percentage of metallic copper in the original mixture? ANS:::: Molar mass of Cu = 63.5 g mol-1 (in some books, 64 g mol-1) Molar mass of CuO = 63.5 + 16 = 79.5 g mol-1 The mixture contains Cu and CuO. Cu is stable when heated in a stream of hydrogen. CuO + H2 → Cu + H2O Let the amount of CuO in the mixture be y mol. No. of moles of Cu formed = y mol Mass of CuO reacted = y x 79.5 = 79.5y g Mass of Cu formed = y x 63.5 = 63.5y g 79.5y - 63.5y = 0.92 - 0.752 16y = 0.168 y = 0.0105 No. of moles of CuO in the mixture = 0.0105 mol Mass of CuO in the mixture = 0.0105 x 79.5 = 0.835 g Percentage by mass of CuO in the mixture = (0.835/0.92) x 100% = 90.76% Percentage by mass of Cu in the mixture = 1 - 90.76% = 9.24% 5 是好明 79.5y - 63.5y = 0.92 - 0.752 16y = 0.168 y = 0.0105 呢個 79.5y - 63.5y 是咩來的 0.92 - 0.752 是咩來的 為何他門相等 咁開頭ge mixture 是咪 Cu+CuO residue 是咪 Cu 同埋reaction 之後ge Cu 咁點解H20 吾計落下residue 到ge??
最佳解答:
CuO(s) + H2-(g) → Cu(s) + H2O(g) 由於 reaction 在加熱之下進行，所以生成的 H2O 變成蒸氣走晒。 假設原來在 mixture 中的 CuO 是 y mol（79.5y g）。 則 Cu produced in reaction 亦是 y mol（63.5y g）。 (Mass of CuO in mixture) - (Mass of Cu produced) = (79.5y - 63.5y) g 但 (Mass of CuO in mixture) - (Mass of Cu produced) = (Mass of CuO in mixture + Mass of Cu in mixture) - (Mass of Cu produced + Mass of Cu in mixture) = (0.92 - 0.752) g 所以，79.5y - 63.5y = 0.92 - 0.752 =
其他解答:
chemistry 問題
發問:
0.92g of a mixture of metallic copper and copper (II) oxide was heated in a stream of hydrogen until reduction was copplete. the residure weighed 0.752 g. what was the percentage of metallic copper in the original mixture? ANS::::Molar mass of Cu = 63.5 g mol-1 (in some books, 64 g mol-1)Molar mass of CuO... 顯示更多 0.92g of a mixture of metallic copper and copper (II) oxide was heated in a stream of hydrogen until reduction was copplete. the residure weighed 0.752 g. what was the percentage of metallic copper in the original mixture? ANS:::: Molar mass of Cu = 63.5 g mol-1 (in some books, 64 g mol-1) Molar mass of CuO = 63.5 + 16 = 79.5 g mol-1 The mixture contains Cu and CuO. Cu is stable when heated in a stream of hydrogen. CuO + H2 → Cu + H2O Let the amount of CuO in the mixture be y mol. No. of moles of Cu formed = y mol Mass of CuO reacted = y x 79.5 = 79.5y g Mass of Cu formed = y x 63.5 = 63.5y g 79.5y - 63.5y = 0.92 - 0.752 16y = 0.168 y = 0.0105 No. of moles of CuO in the mixture = 0.0105 mol Mass of CuO in the mixture = 0.0105 x 79.5 = 0.835 g Percentage by mass of CuO in the mixture = (0.835/0.92) x 100% = 90.76% Percentage by mass of Cu in the mixture = 1 - 90.76% = 9.24% 5 是好明 79.5y - 63.5y = 0.92 - 0.752 16y = 0.168 y = 0.0105 呢個 79.5y - 63.5y 是咩來的 0.92 - 0.752 是咩來的 為何他門相等 咁開頭ge mixture 是咪 Cu+CuO residue 是咪 Cu 同埋reaction 之後ge Cu 咁點解H20 吾計落下residue 到ge??
最佳解答:
CuO(s) + H2-(g) → Cu(s) + H2O(g) 由於 reaction 在加熱之下進行，所以生成的 H2O 變成蒸氣走晒。 假設原來在 mixture 中的 CuO 是 y mol（79.5y g）。 則 Cu produced in reaction 亦是 y mol（63.5y g）。 (Mass of CuO in mixture) - (Mass of Cu produced) = (79.5y - 63.5y) g 但 (Mass of CuO in mixture) - (Mass of Cu produced) = (Mass of CuO in mixture + Mass of Cu in mixture) - (Mass of Cu produced + Mass of Cu in mixture) = (0.92 - 0.752) g 所以，79.5y - 63.5y = 0.92 - 0.752 =
其他解答:
此文章來自奇摩知識+如有不便請留言告知
The first,equation say,1 molCuO can give you 1mol Cu,then he say there have y mol CuO,so you have y mol Cu,because gmol-1xmol=g,also the last have 0.752gCu,79.5y(CuO mass)-63.5y(Cu mass)=oxygen mass=0.92(all mass)-0.752(all Cu mass),also the H2O mass is little,so we don't care for him!!!
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