Question

• 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.​

Answer 1

Answer:

The mass of $CO_{2}$ 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:

$CH_{4} + 2O_{2}$ ⇆ $CO_{2} + 2H_{2}O$

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 = $\frac{44}{16}$ gm of carbon dioxide

∴ 3 moles or (16×3) = 48 gm of methane would give = $\frac{44}{16}$ × 48 gm of carbon dioxide = 132 gm of carbon dioxide.

#SPJ2

Answer 2

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