Question

If the decomposition of 45.987g ammonium nitrate is carried out at 18.0 degrees C and 786.3 mmHg, what volume of dinitrogen monoxide gas will form? The other product of the reaction is gaseous water. What volume of water will form during this reaction?

Answer 1

NH4NO3(s) .............> N2O(g) + 2H2O(g)

From the above balanced equation

1mole of NH4NO3 can produce 1 mole of dinitrogen monoxide and 2moles of H2O

Given 45.987 gm. of NH4NO3

no. of moles of NH4NO3 = (wt/mol.wt) = 45.987/80.043 = 0.574 moles

1 mole of NH4NO3 can produce 1 mole of N2O

then 0.574moles of NH4NO3 can produce ,....................?

   = 0.574 moles of N2O

P = 786.3 mmHg = 1.034 atm.

T = 18 +273 = 291K

n = 0.574 mole

V = ?

According to PV = nRT

R (is gas constant) = 0.0821 lit*atm*mol^-1 K^-1

V = 0.574*0.0821*291/1.034

V = 13.3 lit.

volume of N2O = 13.3 lt

2)

1 mole of NH4NO3 can produce 2 mole of H2O

then 0.574moles of NH4NO3 can produce ,....................?

    = 2*0.574 = 1.15 moles

P = 786.3 mmHg = 1.034 atm.

T = 18 +273 = 291K

n = 1.15 mole

V = ?

According to PV = nRT

V = 1.15*0.0821*291/1.15 = 23.9 lt

volume of gaseous water = 23.9 lit. or 24 lt.