Calculate the volume of CO 2 at 16.5°C and 746 mmHgpressure obtained by treating

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reaction of Na2CO3 with excess of HCl is as follows : Na2CO3 + 2HCl --> 2NaCl + H2O + CO2 so here 1 mole of Na2CO3 = 23*2 + 12+ 16*3 g                                            = 106 g 106 g of Na2CO3 gives 44 g of CO2 so 32.6 gives = (44/106)*32.6 g                      = 13.53 g of CO2 so no of moles of CO2 = 13.53/44                                    = 0.3075 moles so we know that PV = nRT where n = 0.3075 moles and R = universal gas constant = 0.0821 L-atm/K T = 16.5 + 273.15 K     = 289.65 K P = 746 mm Hg we know that 760 mm Hg = 1 atm so 746 mm Hg =( 1/760)*746 atm                        = 0.98157 atm so now putting down the values we get V = nRT/P    = 0.3075*0.0821*289.65/0.98157 L    = 7.4496 L    = 7.45 L so volume of CO2 formed = 7.45 L

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