(#1) Ethanol (C2H5OH ) burns in air: C2H5OH (I) + O2(g) ----> CO2(g) + H2O (I) B
ID: 76499 • Letter: #
Question
(#1)Ethanol (C2H5OH ) burns in air:
C2H5OH (I) + O2(g) ----> CO2(g) + H2O (I)
Balance the equation and determine the volume of air in liters at35.0 degrees C and 790 mmHg required to burn 227 g of ethanol.Assume that air is 21.0 percent O2 by volume.
(#2)
A quantity of 0.225 g of a metal M ( molar mass = 27.0 g/mol )liberated 0.303 L of molecular hydrogen (measured at 17 degrees Cand 741 mmHg) from an excess of hydrochloric acid. Deduce fromthese data the corresponding equation and write formulas for theoxide and sulfate of M .
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Explanation / Answer
(#2) A quantity of 0.225 g of a metal M ( molar mass = 27.0 g/mol )liberated 0.303 L of molecular hydrogen (measured at 17 degrees Cand 741 mmHg) from an excess of hydrochloric acid. Deduce fromthese data the corresponding equation and write formulas for theoxide and sulfate of M . aM + bHCl => aMn+ + (b/2)H2 + bCl- moles of H2 formed: n = PV/(RT) = (741 mmHg *1 atm/ 760 mmHg)*(0.303 L)/(0.0821L*atm/mol/K *290 K) n = 0.012408123 moles H2 formed moles of M consumed 0.225 g *(1mol/27 g) = 0.00833 moles M consumed find stoichiometric ratio: (0.00833 moles M)/(0.012408 moles H2) = 0.67 ~ 2/3 So 3 moles of H2 are formed for every 2 moles of M consumed. That means 2M + 6HCl => 2M3+ + 3 H2 + 6Cl- You know that 2H+ from HCl gets 2 e- to form 1 H2 So 6H+ needs 6e- for 3 H2 That means 6e- from 2M, of 3e- from each M, so M3+ oxide: Because O has -2 charge the oxide isM2O3 2*3+ for 3*2- sulfate also has -2 charge so M2(SO4)3
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