• Calculate the mass co2 in gram produced by the Reaction between 3 mole of CH4 and & 2 mole of Oxygen According to equation - CH4+202 -> CO2 +2H20 Identify the limiting reagend.
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• Calculate the mass co2 in gram produced by the Reaction between 3 mole of CH4 and & 2 mole of Oxygen According to equation - CH4+202 -> CO2 +2H20 Identify the limiting reagend.
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Answer:
The mass of [tex]CO_{2}[/tex] produced in grams by the reaction between 3 moles of methane and 2 moles of oxygen according to the given reaction is 132 grams.
Calculation and Explanation:
Let us write the equation first:
[tex]CH_{4} + 2O_{2}[/tex] ⇆ [tex]CO_{2} + 2H_{2}O[/tex]
1 2 1 2
Thus, we can see that,
1 mole or 16 gm of methane gives 1 mole or 44 gm of carbon dioxide, and,
2 moles or (2×32)= 64 gm of oxygen gives 1 mole or 44 gm of carbon dioxide.
Thus, methane is the limiting reagent here.
What is a limiting reagent?
The limiting reagent in a reaction is that reactant which is totally consumed in the reaction. When the reactant gets consumed, the reaction stops.
Now, we know that,
1 mole or 16 gm of methane gives 1 mole or 44 gm of carbon dioxide.
∴ 1 gm of methane would give = [tex]\frac{44}{16}[/tex] gm of carbon dioxide
∴ 3 moles or (16×3) = 48 gm of methane would give = [tex]\frac{44}{16}[/tex] × 48 gm of carbon dioxide = 132 gm of carbon dioxide.
#SPJ2
44g Of C02
Explanation:
1mole CH4---------»2 mol 02
3 mol CH4-----------»6 mol 02
so Limiting Reagent=O2,(as 6 mol 02 is required to react with 3 mol CH4 but Ony 2 mole Of 02 is avaliable).
2mole Ch4---------»1 mole C02
No Of Mol=Given Mass/Gram Molecular Mass
Gram molecular mass of Co2=44gm
1=Given Mass/44
Mass Of C02=44 Grams
44g of Co2