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1.62g of a metal X (a group 3 element) combine with 1.44g of oxygen to form an oxide. What is the mass of X?A)10.8B)27.0C)52.0D)58.9solution :In group 3 metal, they will lose 3 electrons to obtain the octet structure and oxygen will gain two electrons to obtain the octet. To balance the equation,... 顯示更多 1.62g of a metal X (a group 3 element) combine with 1.44g of oxygen to form an oxide. What is the mass of X? A)10.8 B)27.0 C)52.0 D)58.9 solution : In group 3 metal, they will lose 3 electrons to obtain the octet structure and oxygen will gain two electrons to obtain the octet. To balance the equation, the reaction will be as follows: 2X + 3/2O2 = X2O3 In one oxide molecule, it will contain 3 oxygen atoms and 2 metal X atoms. So, if 1.44g of oxygen is reacted and found in the molecule. the mole will be = 1.44/16 = 0.09. Since the ratio fo O and X in one molcule is 3:2. So, the mol of X willbe = 0.9/ (3:2) = 0.06mol. Since there is 1.62g of X is found in the molecule, the AW of X is 1.62/0.06 = 27. So, the answer will be B (27). 點解係 1.44/16 = 0.09 唔係要1.44/(16*3 )咩??
最佳解答:
Explanation 1: The formula of the compound is X2O3. In one mole of X2O3, there are 2 mol of X atoms and 3 mol of O atoms. Therefore, (No. of moles of X) : (No. of moles of O) = 2 : 3 The ratio 2 : 3 has appeared on the right hand side. It is needless to use the numbers 2 and 3 to calculate the numbers of moles. No. of moles of X = (mass of X)/(molar mass of X) = 1.62/(molar mass of X) No. of moles of O = (mass of X)/(molar mass of O) = 1.44/16 ***** Explanation 2: If there are 1.44 g of O3, The number of moles of O3 = mass/(molar mass of O3 molecules) = 1.44/(16x3) In X2O3, O exists as atoms. There are 1.44 g of O atoms in the compound. No. of moles of O atoms = mass/(molar mass of O atom) = 1.44/16 2008-06-03 01:42:13 補充： To cancio (001 answer): Molar mass is NOT equal to the number of electrons! So, the molar mass of O atom = 16 g mol^-1, but NOT 8, etc. 2008-06-03 01:42:23 補充： Besides, molar mass is the mass of 1 mole of a substance (in g mol^-1), but number of electrons is just a number (no unit). They are two totally DIFFERENT quantaties.
No. of mole = mass / molar mass (Molar mass = no. of electrons of the element) No. of mole of O2 = 1.44 / (8*2) = 1.44/16 (as oxygen has 8 electrons, not 16) If 1.44/8 is wrong, as oxygen should exist in molecule (O2) If 1.44/(8*3) is also wrong, as the oxident is O2, not O3 If 1.44/(16*3) = no. of mole of S3 (as electronic arrangement of S is 2,8,6) I hope the above can help you a lot. ^.^
chem 一問
發問:
1.62g of a metal X (a group 3 element) combine with 1.44g of oxygen to form an oxide. What is the mass of X?A)10.8B)27.0C)52.0D)58.9solution :In group 3 metal, they will lose 3 electrons to obtain the octet structure and oxygen will gain two electrons to obtain the octet. To balance the equation,... 顯示更多 1.62g of a metal X (a group 3 element) combine with 1.44g of oxygen to form an oxide. What is the mass of X? A)10.8 B)27.0 C)52.0 D)58.9 solution : In group 3 metal, they will lose 3 electrons to obtain the octet structure and oxygen will gain two electrons to obtain the octet. To balance the equation, the reaction will be as follows: 2X + 3/2O2 = X2O3 In one oxide molecule, it will contain 3 oxygen atoms and 2 metal X atoms. So, if 1.44g of oxygen is reacted and found in the molecule. the mole will be = 1.44/16 = 0.09. Since the ratio fo O and X in one molcule is 3:2. So, the mol of X willbe = 0.9/ (3:2) = 0.06mol. Since there is 1.62g of X is found in the molecule, the AW of X is 1.62/0.06 = 27. So, the answer will be B (27). 點解係 1.44/16 = 0.09 唔係要1.44/(16*3 )咩??
最佳解答:
Explanation 1: The formula of the compound is X2O3. In one mole of X2O3, there are 2 mol of X atoms and 3 mol of O atoms. Therefore, (No. of moles of X) : (No. of moles of O) = 2 : 3 The ratio 2 : 3 has appeared on the right hand side. It is needless to use the numbers 2 and 3 to calculate the numbers of moles. No. of moles of X = (mass of X)/(molar mass of X) = 1.62/(molar mass of X) No. of moles of O = (mass of X)/(molar mass of O) = 1.44/16 ***** Explanation 2: If there are 1.44 g of O3, The number of moles of O3 = mass/(molar mass of O3 molecules) = 1.44/(16x3) In X2O3, O exists as atoms. There are 1.44 g of O atoms in the compound. No. of moles of O atoms = mass/(molar mass of O atom) = 1.44/16 2008-06-03 01:42:13 補充： To cancio (001 answer): Molar mass is NOT equal to the number of electrons! So, the molar mass of O atom = 16 g mol^-1, but NOT 8, etc. 2008-06-03 01:42:23 補充： Besides, molar mass is the mass of 1 mole of a substance (in g mol^-1), but number of electrons is just a number (no unit). They are two totally DIFFERENT quantaties.
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其他解答:No. of mole = mass / molar mass (Molar mass = no. of electrons of the element) No. of mole of O2 = 1.44 / (8*2) = 1.44/16 (as oxygen has 8 electrons, not 16) If 1.44/8 is wrong, as oxygen should exist in molecule (O2) If 1.44/(8*3) is also wrong, as the oxident is O2, not O3 If 1.44/(16*3) = no. of mole of S3 (as electronic arrangement of S is 2,8,6) I hope the above can help you a lot. ^.^
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