cH4 + 02 =CO2 + H2O
4 gm of methane is burnt with 20gm of oxygen to form CO2 and H2O
a. which one is limiting reactant and why ?
b. calculate the no.of mole of excess reagent
c. calculate the mass of water produced
d. what mass of NaOH is required to consume CO2 produced in the reaction
NaOH + CO2 = Na2CO3 + H2O
Answers
Answer:
81
Explanation:
Answer:
81
Step-by-step explanation:
(a+b)²+a(a+b)+b = 0
a²+2ab+b²+a²+ab+b = 0
2a²+3ab+b+b² = 0
b²+b(3a+1)+2a² = 0
b = \frac{-(3a+1) \pm \sqrt{(3a+1)^2 - 8a^2} }{2}b=2−(3a+1)±(3a+1)2−8a2 [From Sridhar Acharaya's equation]
b = \frac{-(3a+1) \pm \sqrt{9a^2 + 6a + 1 -8a^2} }{2}b=2−(3a+1)±9a2+6a+1−8a2
b = \frac{-(3a+1) \pm \sqrt{ a^2 + 6a + 1} }{y}b=y−(3a+1)±a2+6a+1
b = \frac{-(3a+1) \pm \sqrt{ a(a+6) + 1} }{y}b=y−(3a+1)±a(a+6)+1
As a,b ∈1, a(a + 6) + 1 must be a perfect square
Possible values of a are a = 0 or a = –6
If a = 0, b = –1 or 0
If a = –6, b = 9 or 8
Maximum possible value of b² = 81
Answer:
The reaction is CH 4+2O 2 →CO 2 +2H 2 O.
At room temperature, water will condense to form liquid water.
1 mole of methane reacts with 2 moles of oxygen to form 1 mole of oxygen.
10 ml (0.445 moles) of methane will react with 20 ml (0.89 moles) of oxygen to form 10 ml (0.445 moles) of carbon dioxide.
Hence, the resulting mixture will contain 10 ml of unreacted methane and 10 ml of carbon dioxide.
The total volume of the mixture will be 20 ml.
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