The product in this chemical reaction is Carbon Dioxide or CO2.
This is because it states, "To form," which means that it was produced from the reactants Carbon and Oxygen.
To know what the products are in a statement rather than an equation, words such as produced and formed are used.
Hope this helps!
Answer:
Explanation:
<u>1) Rate law, at a given temperature:</u>
- Since all the data are obtained at the same temperature, the equilibrium constant is the same.
- Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:
r = K [A]ᵃ [B]ᵇ
<u>2) Use the data from the table</u>
- Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s
Divide r₂ by r₁: [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0
- Use the first and second set of data to find the exponent a:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s
Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]
2ᵃ = 2 ⇒ a = 1
<u>3) Write the rate law</u>
This means, that the rate is independent of reactant B and is of first order respect reactant A.
<u>4) Use any set of data to find K</u>
With the first set of data
- r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹
Result: the rate constant is K = 0.167 s⁻¹
From the periodic table:
mass of carbon = 12 grams
mass of hydrogen = 1 gram
mass of chlorine = 35.5 grams
Therefore,
molar mass of CH2Cl2 = 12 + 2(1) + 2(35.5) = 85 grams
number of moles = mass / molar mass
number of moles of CH2Cl2 = 66.05 / 85 = 0.777 moles
One mole of CH2Cl2 contains two moles of Cl and each chlorine mole has Avogadro's number of atoms in it.
Therefore,
number of chlorine atoms in 0.777 moles of CH2Cl2 can be calculated as follows:
number of atoms = 0.777 * 2 * 6.022 * 10^23 = 9.358 * 10^23 atoms
Now, we will take log base 10 for this number:
log (9.358 * 10^23) = 23.97119
Answer:
2 mol H₂O
Explanation:
With the reaction,
- 2H₂(g) + O₂(g) → 2 H₂O(g)
1.55 moles of O₂ would react completely with ( 2*1.55 ) 3.1 moles of H₂. There are not as many moles of H₂, thus H₂ is the limiting reactant.
Now we <u>calculate the moles of H₂O produced</u>, <em>starting from the moles of limiting reactant</em>:
- 2.00 mol H₂ *
= 2 mol H₂O
