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(20分~緊急)3條有關chem計mole既問題~有心人請回答~thx~!
1)Complete combustion of 1.48 g of an organic compound Y gave 2.64 g of carbon dioxide and 1.08 g of water as the only products. If the relative molecular mass of Y is equal to 74, what is the molecular formula of Y? (Relative atomic masses: C = 12.0; H = 1.0; O = 16.0)2)What mass of iron(III) oxide is required... 顯示更多 1)Complete combustion of 1.48 g of an organic compound Y gave 2.64 g of carbon dioxide and 1.08 g of water as the only products. If the relative molecular mass of Y is equal to 74, what is the molecular formula of Y? (Relative atomic masses: C = 12.0; H = 1.0; O = 16.0) 2)What mass of iron(III) oxide is required to obtain 0.5 mole of iron by carbon reduction? (Relative atomic masses: Fe = 55.8; O = 16.0; C = 12.0) 3)How many grams of carbon is required to reduce 11.32 g of copper(II) oxide completely? (Relative atomic masses: Cu = 63.5; O = 16.0) 請有solution +解釋~ (因為唔想只係抄左就算~ 想學習如何計算~有心人請回答~辛苦哂!)
最佳解答:
mass of carbon in the sample = (% by mass of C in CO2)*(mass of CO2 formed) = (12/44)(2.64) = 0.72 g mass of hydrogen in the sample = (2/18)(1.08) = 0.12 g since the sample is burnt in excess oxygen, we must use the method below to find the mass of O mass of O = (1.48-0.72-0.12) = 0.64 g Carbon Hydrogen oxygen mass 0.72 0.12 0.64 No. of moles 0.06 0.12 0.04 Divided by smallest 1.5 3 1 Ratio 3 6 2 (3*12+6*1+2+16)n=74 n=1 so molecular formula of Y is C3H6O2 2Fe2O3 + 3C---->4Fe + 3CO2 (no. of moles of iron(III) oxide required)/(no of moles of Fe formed) =2/4 no. of moles of iron(III) oxide required =(1/2)(0.5)=0.25 mass of iron(III) oxide required =0.25(55.8*2+16*3)=39.9g 2CuO+C---->2Cu + CO2 no. of moles of C =(1/2)(no. of moles of CuO) =0.5*11.32/(63.5+16) =0.071195 mass of carbon required =0.071195*12 =0.854g
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(20分~緊急)3條有關chem計mole既問題~有心人請回答~thx~!
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發問:1)Complete combustion of 1.48 g of an organic compound Y gave 2.64 g of carbon dioxide and 1.08 g of water as the only products. If the relative molecular mass of Y is equal to 74, what is the molecular formula of Y? (Relative atomic masses: C = 12.0; H = 1.0; O = 16.0)2)What mass of iron(III) oxide is required... 顯示更多 1)Complete combustion of 1.48 g of an organic compound Y gave 2.64 g of carbon dioxide and 1.08 g of water as the only products. If the relative molecular mass of Y is equal to 74, what is the molecular formula of Y? (Relative atomic masses: C = 12.0; H = 1.0; O = 16.0) 2)What mass of iron(III) oxide is required to obtain 0.5 mole of iron by carbon reduction? (Relative atomic masses: Fe = 55.8; O = 16.0; C = 12.0) 3)How many grams of carbon is required to reduce 11.32 g of copper(II) oxide completely? (Relative atomic masses: Cu = 63.5; O = 16.0) 請有solution +解釋~ (因為唔想只係抄左就算~ 想學習如何計算~有心人請回答~辛苦哂!)
最佳解答:
mass of carbon in the sample = (% by mass of C in CO2)*(mass of CO2 formed) = (12/44)(2.64) = 0.72 g mass of hydrogen in the sample = (2/18)(1.08) = 0.12 g since the sample is burnt in excess oxygen, we must use the method below to find the mass of O mass of O = (1.48-0.72-0.12) = 0.64 g Carbon Hydrogen oxygen mass 0.72 0.12 0.64 No. of moles 0.06 0.12 0.04 Divided by smallest 1.5 3 1 Ratio 3 6 2 (3*12+6*1+2+16)n=74 n=1 so molecular formula of Y is C3H6O2 2Fe2O3 + 3C---->4Fe + 3CO2 (no. of moles of iron(III) oxide required)/(no of moles of Fe formed) =2/4 no. of moles of iron(III) oxide required =(1/2)(0.5)=0.25 mass of iron(III) oxide required =0.25(55.8*2+16*3)=39.9g 2CuO+C---->2Cu + CO2 no. of moles of C =(1/2)(no. of moles of CuO) =0.5*11.32/(63.5+16) =0.071195 mass of carbon required =0.071195*12 =0.854g
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