Given the model from the question,
- The products are: N₂, H₂O and H₂
- The reactants are: H₂ and NO
- The limiting reactant is H₂
- The balanced equation is: 3H₂ + 2NO —> N₂ + 2H₂O + H₂
<h3>Balanced equation </h3>
From the model given, we obtained the ffolowing
- Red => Oxygen
- Blue => Nitrogen
- White => Hydrogen
Thus, we can write the balanced equation as follow:
3H₂ + 2NO —> N₂ + 2H₂O + H₂
From the balanced equation above,
- Reactants: H₂ and NO
- Product: N₂, H₂O and H₂
<h3>How to determine the limiting reactant</h3>
3H₂ + 2NO —> N₂ + 2H₂O + H₂
From the balanced equation above,
3 moles of H₂ reacted with 2 moles of NO.
Therefore,
5 moles of H₂ will react with = (5 × 2) / 3 = 3.33 moles of NO
From the calculation made above, we can see that only 3.33 moles of NO out of 4 moles given are required to react completely with 5 moles of H₂.
Thus, H₂ is the limiting reactant
Learn more about stoichiometry:
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It is true yes :) happy to help
M1v1=m2v2
m2=(m1v1)/v2
Where m is the molarities and v is the volumes
<span>m2=(25.0*0.500)/53.5
m2=12.5/53.5
m2=0.2336
by rounding off:
m2=0.234 M
so the answer is C: 0.234 M</span>
Solid water is less dense than liquid water.
