Question

Ammonia gas is synthesized by combining hydrogen and nitrogen. A. To produce 562g of NH3 what volume of H2 gas is required at 56 degrees celcius and 745 mmHg? B. To produce 562g of NH3 what volume of N2 is required if introduced at 29 celcius and 745mmHg

Answer 1

m = 562 g of NH3

V of H2 gas reqruied at

T = 56C = 56+273 = 329K

P = 745 mm Hg = 745/760 atm = 0.98 atm

calculate mol needed

N2 + 3H2 = 2NH3

then

mol of NH3 = mass/MW = 562/17 = 33.0588 mol of NH3

then

ratio is 3 mol of H2 : 2 molo fNH3

then

3/2 is the ratio

49.5882 mol of H2

change to volume with ideal gas law

PV = nRt

V = nRT/(P)

V = (49.5882)(0.082)(329)/(0.98) = 1365.09 L

b)

m = 562 g of NH3

mol of NH3 = mass/MW = 562/17 = 33.0588 molf NH3

mol of N2 needed

1 mol N2 : 2 NH3

1:2

ratio then

33.0588/2 = 16.53 mol of N2

PV = nRT

V = nRT/P

T = 29 = 29+273 = 302

V = (16.53)(0.082)(302)/(0.98) = 417.70 L