An acetylene tank for an oxacetylene welding torch provides9341L of acetylene ga

Question

An acetylene tank for an oxacetylene welding torch provides9341L of acetylene gas, C2H2 at0oC and 745 torr. how many tanks of oxygen, eachproviding 7.00x103L of O2 at 0oCand 745 torr, will be required to burn all of the acetylene? Theequation is given below 2C2H2 (g) + 5O2 (g)-------> 4CO2 (g) + H2O (g) An acetylene tank for an oxacetylene welding torch provides9341L of acetylene gas, C2H2 at0oC and 745 torr. how many tanks of oxygen, eachproviding 7.00x103L of O2 at 0oCand 745 torr, will be required to burn all of the acetylene? Theequation is given below 2C2H2 (g) + 5O2 (g)-------> 4CO2 (g) + H2O (g)

Explanation / Answer

2C2H2 (g) + 5O2 (g)-------> 4CO2 (g) + H2O (g) 2C2H2 (g) + 5O2 (g)-------> 4CO2 (g) + H2O (g) We know that PV = nRT Where P = pressure= 745 torr       Since 760 torr= 1 atm                               = 745 / 760 atm                               = 0.9802 atm T = Temperature = 0 o C = 273 K V = Volume = 9341 L n = No . of moles = ? R = gas constant = 0.0821 L atm . mol - K Plug the values we get , n = PV / RT                                       = 408.6 moles no . of mole of each tank of oxygen , n' = PV / RT                                                            = 0.9802 atm * 7000 L / (0.0821 L atm / mol - k * 273 K )                                                            = 306.13 moles 2 moles of C2H2 require 5 moles of O2 to complete thereaction 408.6 moles of C2H2 require Xmoles of O2 to complete thereaction X = ( 5 * 408.6 ) / 2     = 1021.5 moles So no . of tanks of O2 gas required is , N = X / n'                                                                = 3.336 tanks                                                                 

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