Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane: 2 nh3 (g) + 3 o2 (g) + 2 ch4 (g) → 2 hcn (g) + 6 h2o (g) if 5.00 x 103 kg each of nh3, o2, and ch4 are reacted, what mass of hcn and of h 2o will be produced, assuming 100% yield?

mass of HCN=7934.14Kg, mass of H2O=2812.5

Explanation:

Since there is 100% yield,it means all the reactants are converted in to products

2 NH3 (g) + 3 O2 (g) + 2 CH4 (g) → 2 HCN (g) + 6 H2O (g)

molar mass of O2=16

molar mass of HCN=27.02

molar mass of H2O=18

Applying mole ratio

no of mole =mass/molar mass

since ammonia(NH3)  is converted in to Hydrogen cyanide(HCN)

no of mole of ammonia =no of mole of hydeogen cyanide

1:1

mass of NH3/molar mass of  NH3=mass of HCN/molar mas of HCN

by making :mass of HCN the subject,

mass of HCN=5000×27.02/17.031

Mass of HCN=7934.14kg

applying the same mole ratio

Oxygen is converted in to water

no of mole of O2=2×no of mole of H2O

Mass of O2/molar mass of O2=2×mass of H2O/molar mass of H2O

by making mass of H2O the subject

mass of H2O=5000×18/32

=2815.5kg

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm