A link is a part of a chain and an atom is a part of a molecule. It is also a part of a compound. Your answer should be molecule and/or compound
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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Answer:
D. The piece of bread travels with straight- line horizontal acceleration.
Explanation
Answer:
15.0 L
Explanation:
To find the volume, you need to use the Ideal Gas Law:
PV = nRT
In this equation,
-----> P = pressure (mmHg)
-----> V = volume (L)
-----> n = moles
-----> R = Ideal Gas constant (62.36 L*mmHg/mol*K)
-----> T = temperature (K)
To calculate the volume, you need to (1) convert grams C₄H₁₀ to moles (via the molar mass), then (2) convert the temperature from Celsius to Kelvin, and then (3) calculate the volume (via the Ideal Gas Law).
Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)
Molar Mass (C₄H₁₀): 58.124 g/mol
32 grams C₄H₁₀ 1 moles
------------------------- x ----------------------- = 0.551 moles C₄H₁₀
58.124 grams
P = 728 mmHg R = 62.36 L*mmHg/mol*K
V = ? L T = 45.0 °C + 273.15 = 318.15 K
n = 0.551 moles
PV = nRT
(728 mmHg)V = (0.551 moles)(62.36 L*mmHg/mol*K)(318.15 K)
(728 mmHg)V = 10922.7632
V = 15.0 L
Reactives
-> Products
CuO
and water are products.
I
found this reaction which has CuO and water as products: decomposition of
Cu(OH)2.
Cu(OH)2
-> CuO + H2O
Stoichiometry calculus involve the mole
proportions you can see in the reaction: When 1 mole of Cu(OH)2 reacts, 1 mole of
CuO and 1 mole of H2O are formed.
Considering
the molar masses:
Cu(OH)2
= 83.56 g/mol
CuO
= 79.545 g/mol
H2O
= 18.015 g/mol
Then:
When 83.56 g of Cu(OH)2 react, 79.545 g of CuO and 18.015 g H2O are formed.
You
should use that numbers in the rule of three:
79.545
g CuO __________18.015 g water
3.327
g CuO__________ x =3.327*18.015 /79.545 g water
x= 0.7535 g water
