Question

Aqueous nitric acid reacts with solid sodium carbonate to produce aqueous sodium nitrate, carbon dioxide, and water. If 10.0 mL of a 1.00 M nitric acid solution reacts with 5.00 g of sodium carbonate, what volume of dry carbon dioxide will be produced at a temperature of 32.5 degrees Celcius at a pressure of 0.96 atm?

Answer 1

2HNO3(aq) + Na2Co3(aq) -----------------> 2NaNO3(aq) + H2O + CO2(g)

no of moles of HNO3        =   molarity *volume in L

                                          = 1*0.01   = 0.01 moles

no of moles of Na2CO3   = W/G.M.Wt

                                        = 5/106   = 0.047 moles

1 moles of Na2Co3 react with 2 moles of HNO3

0.047 moles of Na2CO3 react with = 2*0.047/1 = 0.094 moles of HNO3 is required

   HNO3 is limiting reactant

2 moles of HNO3 react with Na2Co3 to gives 1 mole of Co2

0.01 moles of HNO3 react with Na2Co3 to gives = 0.01*1/2    = 0.005 moles of CO2

PV = nRT

T = 32.5+273 = 305.5K

P   = 0.96atm

n = 0.005 moles

PV = nRT

V = nRT/P

   = 0.005*0.0821*305.5/0.96   = 0.13L >>>>>answer

volume of Co2 = 0.13L