determine mass of water formed when 12.5 L NH3 (at298K and 1.50atm) is reacted with 18.9L of O2 (at 323K and 1.1atm)

The mass of water formed is

calculation

Use the ideal gas equation to calculate the moles of NH3 and O2

that is Pv = n RT

where; P = pressure,

V = volume,

n = number of moles,

R=gas constant = 0.0821 l. atm / mol. K

make n the formula of the subject by diving both side by RT

n = PV / RT

The moles of NH3

n = (1.50 atm x 12.5 L) / (0.0821 L. atm / mol. k x 298 K) = 0.766 moles

The moles of O2

= (1.1 atm x 18.9 L) / (0.0821 L. atm / mol. k x 323 K) = 0.784 moles

write the reaction between NH3 and O2

4 NH3 + 5 O2 →4 No + 6H2O

from equation above 0.766 moles of NH3 reacted to produce

0.766 x 6/4 = 1.149 moles of H2O

0.784 moles of O2 reacted to produce 0.784 x 6/5=0.9408 moles of H20

since O2 is totally consumed, O2 is the limiting reagent and therefore the moles of H2O produced = 0.9408 moles

mass of H2O = moles x molar mass

from periodic table the molar mass of H2O = (1 x2) + 16 = 18 g/mol

mass = 18 g/mol x 0.9408 moles = 16.93 grams