Question

How many milliliters of water vapor, H₂O_(g), measured at 318 ^oC and 735 torr, are formed when 33.6 mL of NH₃ at 825 torr and 127 ^oC reacts with oxygen according to the following equation:

4NH_3(g) + 3O_2(g) --> 2N_2(g) + 6H₂O_(g)

Answer 1

825 torr = 1.08553 atm

127⁰C = 400.15 K

33.6 ml = 0.0336 L

Use ideal gas equation for calculation of mole of NH₃

Ideal gas equation

PV = nRT             where, P = atm pressure= 825torr = 1.08553 atm,

V = volume in Liter = 33.6 ml = 0.0336 L

n = number of mole = ?

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 127⁰C = 273.15+ 127 = 400.15 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (1.085530.0336)/(0.08205400.15) = 0.0011109 mole

4NH₃(g) + 3O₂(g) --> 2N₂(g) + 6H₂O(g)

According to reaction 4 mole of ammonia produce 6 mole of water then 0.0011109 mole ammonia produce

0.0011109 6 / 4 = 0.001666 mole of water vapour

We know that PV = nRT

V = nRT/P

n = 0.001666 mole, T = 318⁰C = 591.15K, P= 735 torr = 0.967105 atm, R = 0.0821 L atm mol^-1 K^-1 ( R = gas constant)

Substitute these value in above equation.

V = 0.0016660.08205591.15/0.967105 = 0.0835744 L = 83.5744 ml

Volume of water vapour is 83.5744 ml