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 :

<span>0.0687 m
The balanced equation is
BaCl2 + Na2SO4 ==> BaSO4 + 2 NaCl
Looking at the equation, it indicates that there's a 1 to 1 ratio of BaCl2 and Na2SO4 in the reaction. So the number of moles of each will be equal. Now calculate the number of moles of Na2SO4 we had. Start by looking up atomic weights.
Atomic weight sodium = 22.989769
Atomic weight sulfur = 32.065
Atomic weight oxygen = 15.999
Molar mass Na2SO4 = 2 * 22.989769 + 32.065 + 4 * 15.999 = 142.040538 g/mol
Moles Na2SO4 = 0.554 g / 142.040538 g/mol = 0.003900295 mol
Molarity is defined as moles per liter, so let's do the division.
0.003900295 mol / 0.0568 l = 0.068667165 mol/l = 0.068667165 m
Rounding to 3 significant figures gives 0.0687 m</span>
Answer:
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