1) If 1×10 3 of Cytosine is submitted to combustionanalysis, (a) Howmany moles o

Question

1) If 1×103 ofCytosine is submitted to combustionanalysis,          (a) Howmany moles of CO2 would be formed?          (b) How manymoles of H2O would be formed? 2) How many moles are in a gram of each of the followingsubstances?      Si , Cl2 , Au,NH3 1) If 1×103 ofCytosine is submitted to combustionanalysis,          (a) Howmany moles of CO2 would be formed?          (b) How manymoles of H2O would be formed? 2) How many moles are in a gram of each of the followingsubstances?      Si , Cl2 , Au,NH3

Explanation / Answer

(1) 2C4H5N3O + 1202 ---> 8 CO2 + 3 N2+ 10 H20 I think that the given value is no. of moles 2 moles of Cytosine on combustion gives8 moles of O2 1 * 10^-3 moles of Cytosine oncombustion gives X moles of O2 X = ( 8 * 10^-3 ) / 2     = 4 * 10^-3 moles of O2 2 moles of Cytosine on combustiongives 10 moles of H2O 1 * 10^-3 moles of Cytosine oncombustion gives Y moles of H2O X = ( 10 * 10^-3 ) / 2     = 5 * 10^-3 moles of H2O (2) Molar mass of Si is 28g No. of moles = mass / Molar mass                     = 1 / 28                     = 0.0357 moles Molar mass of Cl 2 is 2 * 35.5 = 71 g No. of moles = mass / Molar mass                     = 1 / 71                     = 0.014 moles Molar mass of Au is 197 g No. of moles = mass / Molar mass                     = 1 / 197                     = 0.005076 moles Molar mass of NH3 = 14 + 3 * 1= 17 g No. of moles = mass / Molar mass                     = 1 / 17                     = 0.0588 moles 2 moles of Cytosine on combustiongives 10 moles of H2O 1 * 10^-3 moles of Cytosine oncombustion gives Y moles of H2O X = ( 10 * 10^-3 ) / 2     = 5 * 10^-3 moles of H2O (2) Molar mass of Si is 28g No. of moles = mass / Molar mass                     = 1 / 28                     = 0.0357 moles Molar mass of Cl 2 is 2 * 35.5 = 71 g No. of moles = mass / Molar mass                     = 1 / 71                     = 0.014 moles Molar mass of Au is 197 g No. of moles = mass / Molar mass                     = 1 / 197                     = 0.005076 moles Molar mass of NH3 = 14 + 3 * 1= 17 g No. of moles = mass / Molar mass                     = 1 / 17                     = 0.0588 moles No. of moles = mass / Molar mass                     = 1 / 71                     = 0.014 moles Molar mass of Au is 197 g No. of moles = mass / Molar mass                     = 1 / 197                     = 0.005076 moles Molar mass of NH3 = 14 + 3 * 1= 17 g No. of moles = mass / Molar mass                     = 1 / 17                     = 0.0588 moles No. of moles = mass / Molar mass                     = 1 / 197                     = 0.005076 moles Molar mass of NH3 = 14 + 3 * 1= 17 g No. of moles = mass / Molar mass                     = 1 / 17                     = 0.0588 moles No. of moles = mass / Molar mass                     = 1 / 17                     = 0.0588 moles

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