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a. volume of NO : 41.785 L
b. mass of H2O : 18 g
c. volume of O2 : 9.52 L
<h3>Further explanation</h3>Given
Reaction
4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)
Required
a. volume of NO
b. mass of H2O
c. volume of O2
Solution
Assume reactants at STP(0 C, 1 atm)
Products at 1000 C (1273 K)and 1 atm
a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :
volume NO at 1273 K and 1 atm
b. 15 L NH3 at STP ( 1mol = 22.4 L)
mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :
mass H2O(MW = 18 g/mol) :
c. mol NO at 1273 K and 1 atm :
mol ratio of NO : O2 = 4 : 5, so mol O2 :
Volume O2 at STP :