Answer:
The answer to your question is a) N₂ b) 3.04 g of NH₃
Explanation:
Data
mass of H₂ = 2.5 g
mass of N₂ = 2.5 g
molar mass H₂ = 2.02 g
molar mass of N₂ = 28.02 g
molar mass of NH₃ = 17.04 g
Balanced chemical reaction
3H₂ + 1 N₂ ⇒ 2NH₃
A)
Calculate the theoretical yield 3H₂ / N₂ = 3(2.02) / 28.02 = 0.22
Calculate the experimental yield H₂/N₂ = 2.5/2.5 = 1
Conclusion
The limiting reactant is N₂ (nitrogen) because the experimental proportion was higher than the theoretical proportion.
B)
28.02 g of N₂ -------------------- (2 x 17.04) g of NH₃
2.5 g of N₂ -------------------- x
x = (2.5 x 2 x 17.04) / 28.02
x = 85.2 / 28.02
x = 3.04 g of NH₃
Answer:
Reduction
Explanation:
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Mn⁺⁷ +3e⁻ → Mn⁴⁺
Mn gets three electrons , its oxidation state reduced from +7 to +4 so Mn gets reduced.
Examples:
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
Answer: C (The first species to populate an uninhabited area)
Explanation:
C. 96.82 kPa because to find the amount of pressure the air is causing you need to subtract the amount of pressure the water vapor is causing because the only two gasses making up the air in the pool area are air and water vapor.
<span>Stoichiometry deals with the quantitative measurement of reactants and products in a chemical reaction. Let suppose you are given with following reaction;
A + 2 B </span>→ 3 C
According to this reaction 1 mole of A reacts with 2 moles of B to produce 3 moles of C. Now using the concept of mole one can easily measure the amount of reactants reacted and the amount of product formed, as...
1 Mole Exactly equals 6.022 × 10²³ particles
1 Mole of Gas (at STP) exactly occupies 22.4 L Volume
1 Mole of any compound exactly equals the molar mass in grams
Therefore, <span>Stoichiometry is very helpful in quantitative analysis.</span>
