標題:
數條chem計數問題 高手幫幫忙! (20 marks)
1. amminia is produced :N2 (g) + 3H2 (g) --> 2NH3(g)a. calculate the theoretical yield of ammonia produced from 72.8g of nitrogen2. a compound X contains 54.5% carbon 9.10% hydrogen and 36.4% oxygen its relative molecular mass is 88.0 (relative atomic masses: H =1.0 C=12.0 O=16.0) a. find its... 顯示更多 1. amminia is produced : N2 (g) + 3H2 (g) --> 2NH3(g) a. calculate the theoretical yield of ammonia produced from 72.8g of nitrogen 2. a compound X contains 54.5% carbon 9.10% hydrogen and 36.4% oxygen its relative molecular mass is 88.0 (relative atomic masses: H =1.0 C=12.0 O=16.0) a. find its empirical formula b. find its molecular formula
最佳解答:
1) N2(g) + 3H2(g) --> 2NH3(g) From the equation, we can describe in words that: 1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia gas. So suppose there are excess supply of hydrogen gas, i.e. nitrogen gas is the limiting reactant: Molar mass of nitrogen (N2) = 2 x 14 = 28g So mo. of moles of nitrogen gas in 72.8g = 72.8/28 = 2.6 moles So to speak, if the reaction is complete, a maximum of 2.6 x 2 = 5.2 moles of ammonia gas will be produced. And this amount is equal to 5.2 x (14 + 3 x 1) = 88.4g or 5.2 x 24 = 124.8 dm^3 measured under room temperature and pressure. 2) a) Suppose there is 100g of such compound. Then, mass of C = 54.5g = 54.5/12 = 4.54 moles mass of H = 9.1g = 9.1/1 = 9.1 moles mass of O = 36.4g = 36.4/12 = 2.275 moles Thus C : H : O = 4.54 : 9.1 : 2.275 and approximates to C : H : O = 2 : 4 : 1 So the empirical formula is C2H4O b) Its empirical mass = 2 x 12 + 4 x 1 + 16 = 44 Hence its molecular mass is twice its empirical mass and then the molecular formula is double of the empirical formula, i.e. C4H8O2
其他解答:
1. In this reaction, we assumed all the nitrogen used can be 100% converted to ammonia. Since we have 72.8 gm of nitrogen molecule, the mole of nitorgen atom is 72.7/ 28 x 2 = 5.2 mol. THis 5.2 mol of ammonia will be completely converted to Ammonia (MW = 17). So, the weight of ammonia formed = 5.2 x 17 = 88.4gm. 2a. The empirical formula is C= 54.5/12; H = 9.1/1; O=36.4/16 = 4.54: 9.1:2.275 = 2:4:1 So, the empirical formula should be C2H4O. 2b. The molecular formula is C: 54.5%x88/12 = 4; H: 9.1%x88/1 = 8 and O: 36.4%x88/16 = 2. So, the molecular formula is C4H8O2.|||||1a. N2(g) + 3H2(g) --> 2NH3(g) mole ratio of N2 : H2 : NH3 = 1 :3 : 2 no. of mole of 72.8g of nitrogen = mass / molar mass = 72.8/14= 5.2mol According to the mole ratio of the formula. N2 : NH3 = 1 : 2 Therefore the theorectical yeild of NH3 is 5.2 x 2 = 10.4mol 10.4 mol of NH3: mole x molar mass= mass= 10.4x(14+1X3)= 176.8g 2a. Epirical formula is C= 54.5% H= 9.1 % O= 36.4% Let the compound is 100g C= 54.5g; H=9.1g; O=36.4g C : H : O mol= 54.5g/12; 9.1g/1 ; 36.4g/16 4.54 : 9.1 :2.2 mol ratio=4.54/2.2: 9.1/2.2: 36.4/2.2 2.06 : 4.14 : 1 therefore the empirical formula is C: H: O = 2:4:1= C2H4O b. molecular mass is 88.0 molecular mass of the empirical formula is (12x2 +1x4 + 16) = 44 88/ 44 =2 the molecular formula of the compound X is (C2H4O) 2 = C4H8O2|||||Sorry,I only know Q2... 2a. You should assume that there is 100g of compound.That means there are 54.5 g of carbon,9.1g of hydrogen and etc. C H O Mass: 54.5 9.1 36.4 Molar mass: 12 1 16 Number of mole: 4.542 9.1 2.275 Divided by the smallest number of mole Mole ratio: 1.996(~2) 4 1 So,the empirical formula: C2H4O b. Let the molecular formula of the compound be (C2H4O)n,where n is a whole number. (C2H4O)n=88 (12X2+1X4+16)n=88 44n=88 n=2 So,the molecular formula of compound X is C4H8O2. Hope I can help u~~~
數條chem計數問題 高手幫幫忙! (20 marks)
此文章來自奇摩知識+如有不便請留言告知
發問:1. amminia is produced :N2 (g) + 3H2 (g) --> 2NH3(g)a. calculate the theoretical yield of ammonia produced from 72.8g of nitrogen2. a compound X contains 54.5% carbon 9.10% hydrogen and 36.4% oxygen its relative molecular mass is 88.0 (relative atomic masses: H =1.0 C=12.0 O=16.0) a. find its... 顯示更多 1. amminia is produced : N2 (g) + 3H2 (g) --> 2NH3(g) a. calculate the theoretical yield of ammonia produced from 72.8g of nitrogen 2. a compound X contains 54.5% carbon 9.10% hydrogen and 36.4% oxygen its relative molecular mass is 88.0 (relative atomic masses: H =1.0 C=12.0 O=16.0) a. find its empirical formula b. find its molecular formula
最佳解答:
1) N2(g) + 3H2(g) --> 2NH3(g) From the equation, we can describe in words that: 1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia gas. So suppose there are excess supply of hydrogen gas, i.e. nitrogen gas is the limiting reactant: Molar mass of nitrogen (N2) = 2 x 14 = 28g So mo. of moles of nitrogen gas in 72.8g = 72.8/28 = 2.6 moles So to speak, if the reaction is complete, a maximum of 2.6 x 2 = 5.2 moles of ammonia gas will be produced. And this amount is equal to 5.2 x (14 + 3 x 1) = 88.4g or 5.2 x 24 = 124.8 dm^3 measured under room temperature and pressure. 2) a) Suppose there is 100g of such compound. Then, mass of C = 54.5g = 54.5/12 = 4.54 moles mass of H = 9.1g = 9.1/1 = 9.1 moles mass of O = 36.4g = 36.4/12 = 2.275 moles Thus C : H : O = 4.54 : 9.1 : 2.275 and approximates to C : H : O = 2 : 4 : 1 So the empirical formula is C2H4O b) Its empirical mass = 2 x 12 + 4 x 1 + 16 = 44 Hence its molecular mass is twice its empirical mass and then the molecular formula is double of the empirical formula, i.e. C4H8O2
其他解答:
1. In this reaction, we assumed all the nitrogen used can be 100% converted to ammonia. Since we have 72.8 gm of nitrogen molecule, the mole of nitorgen atom is 72.7/ 28 x 2 = 5.2 mol. THis 5.2 mol of ammonia will be completely converted to Ammonia (MW = 17). So, the weight of ammonia formed = 5.2 x 17 = 88.4gm. 2a. The empirical formula is C= 54.5/12; H = 9.1/1; O=36.4/16 = 4.54: 9.1:2.275 = 2:4:1 So, the empirical formula should be C2H4O. 2b. The molecular formula is C: 54.5%x88/12 = 4; H: 9.1%x88/1 = 8 and O: 36.4%x88/16 = 2. So, the molecular formula is C4H8O2.|||||1a. N2(g) + 3H2(g) --> 2NH3(g) mole ratio of N2 : H2 : NH3 = 1 :3 : 2 no. of mole of 72.8g of nitrogen = mass / molar mass = 72.8/14= 5.2mol According to the mole ratio of the formula. N2 : NH3 = 1 : 2 Therefore the theorectical yeild of NH3 is 5.2 x 2 = 10.4mol 10.4 mol of NH3: mole x molar mass= mass= 10.4x(14+1X3)= 176.8g 2a. Epirical formula is C= 54.5% H= 9.1 % O= 36.4% Let the compound is 100g C= 54.5g; H=9.1g; O=36.4g C : H : O mol= 54.5g/12; 9.1g/1 ; 36.4g/16 4.54 : 9.1 :2.2 mol ratio=4.54/2.2: 9.1/2.2: 36.4/2.2 2.06 : 4.14 : 1 therefore the empirical formula is C: H: O = 2:4:1= C2H4O b. molecular mass is 88.0 molecular mass of the empirical formula is (12x2 +1x4 + 16) = 44 88/ 44 =2 the molecular formula of the compound X is (C2H4O) 2 = C4H8O2|||||Sorry,I only know Q2... 2a. You should assume that there is 100g of compound.That means there are 54.5 g of carbon,9.1g of hydrogen and etc. C H O Mass: 54.5 9.1 36.4 Molar mass: 12 1 16 Number of mole: 4.542 9.1 2.275 Divided by the smallest number of mole Mole ratio: 1.996(~2) 4 1 So,the empirical formula: C2H4O b. Let the molecular formula of the compound be (C2H4O)n,where n is a whole number. (C2H4O)n=88 (12X2+1X4+16)n=88 44n=88 n=2 So,the molecular formula of compound X is C4H8O2. Hope I can help u~~~
文章標籤
全站熱搜
留言列表
