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 :

Answer:
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Explanation:
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Answer:Molarity
Explanation:M stand for molarity
Answer:
146 g
Explanation:
Step 1. Calculate the <em>molar mass</em> of NaNO₃
Na = 22.99
N = 14.01
3O = 3 × 16.00 = 48.00
Total = 85.00 g/mol
Step 2. Calculate the <em>mass</em> of NaNO₃
Mass of NaNO₃ = 1.72 × 85.00/1
Mass of NaNO₃ = 146 g
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