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:
- The molarity of the student's sodium hydroxide solution is 0.0219 M
Explanation:
<u>1) Chemical reaction.</u>
a) Kind of reaction: neutralization
b) General form: acid + base → salt + water
c) Word equation:
- sodium hydroxide + oxalic acid → sodium oxalate + water
d) Chemical equation:
- NaOH + H₂C₂O₄ → Na₂C₂O₄ + H₂O
b) Balanced chemical equation:
- 2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O
<u>2) Mole ratio</u>
- 2mol Na OH : 1 mol H₂C₂O₄ :1 mol Na₂C₂O₄ : 2 mol H₂O
<u>3) Starting amount of oxalic acid</u>
- mass = 28 mg = 0.028 g
- molar mass = 90.03 g/mol
- Convert mass in grams to number of moles, n:
n = mass in grams / molar mass = 0.028 g / 90.03 g/mol = 0.000311 mol
<u>4) Titration</u>
- Volume of base: 28.4 mL = 0.0248 liter
- Concentration of base: x (unknwon)
- Number of moles of acid: 2.52 mol (calculated above)
- Proportion using the theoretical mole ratio (2mol Na OH : 1 mol H₂C₂O₄)
That means that there are 0.000622 moles of NaOH (solute)
<u>5) Molarity of NaOH solution</u>
- M = n / V (liter) = 0.000622 mol / 0.0284 liter = 0.0219 M
That is the correct number using <em>three signficant figures</em>, such as the starting data are reported.
H2SO.Mgslfurmobile phase in this experiment
Amount or merchandise- please give me brainliest please
Answer:
Average atomic mass = 85.557 amu.
Explanation:
Given data:
Percent abundance of Rb-85 = 72.15%
Percent abundance of Rb-87 = 27.85%
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (72.15×85)+(27.85×87) /100
Average atomic mass = 6132.75 + 2422.95 / 100
Average atomic mass = 8555.7 / 100
Average atomic mass = 85.557 amu.
