What is the excess reactant in the combustion of 23 g of methane in the open atmosphere?
1 answer:
Answer : The excess reactant in the combustion of methane in opem atmosphere is molecule.
Solution : Given,
Mass of methane = 23 g
Molar mass of methane = 16.04 g/mole
The Net balanced chemical reaction for combustion of methane is,
First we have to calculate the moles of methane.
=
= 1.434 moles
From the above chemical reaction, we conclude that
1 mole of methane react with the 2 moles of oxygen
and 1.434 moles of methane react to give moles of oxygen
The Moles of oxygen = 2.868 moles
Now we conclude that the moles of oxygen are more than the moles of methane.
Therefore, the excess reactant in the combustion of methane in open atmosphere is molecule.
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Answer:
B. First order, Order with respect to C = 1
Explanation:
The given kinetic data is as follows:
A + B + C → Products
[A]₀ [B]₀ [C]₀ Initial Rate (10⁻³ M/s)
1. 0.4 0.4 0.2 160
2. 0.2 0.4 0.4 80
3. 0.6 0.1 0.2 15
4. 0.2 0.1 0.2 5
5. 0.2 0.2 0.4 20
The rate of the above reaction is given as:
where x, y and z are the order with respect to A, B and C respectively.
k = rate constant
[A], [B], [C] are the concentrations
In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.
Order w.r.t A : Use trials 3 and 4
Order w.r.t B : Use trials 2 and 5
Order w.r.t C : Use trials 1 and 2
we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:
z = 1
Therefore, order w.r.t C = 1