In the reaction ch4+2o2=co2+2h2o how many moles of oxygen are required to burn 16.0 g of methane

#CH_4+2O_2rarrCO_2+2H_2O#

Imole #rarr#2 moles

Convert to grams.

( #A_rC=12,A_rH=1,A_rO=16)#

#[12+(4xx1)]grarr2xx[16+(2xx1)]g#

#16grarr2xx18=36g#

I have assumed the mass of methane to be #5xx10^3g#.

So #1grarr36/16g#

So #5xx10^(3)grarr(36)/(16)xx5xx10^(3)g#

#=11.25xx10^(3)g#

#=11.25kg#

Limiting reagent is the reactant used which produces lesser amount of the product as compared to other reactants if other reactants are used in sufficient amounts. Thus it limits the quantity of the product.

Now, according to your question,

Moles of CH4

= (Mass of CH4 taken)÷(Molecular mass of CH4)

= 8/[12+(4×1)] = 8/16

= 0.5 moles of CH4 are present

Also,

Moles of O2 = (Mass of O2 taken)÷(Molecular mass of O2)

= 16/(16×2) = 16/32

= 0.5 moles of O2 are present

The reaction given is

CH4 + 2O2 => CO2 + 2H2O

Thus,

For 1 mole of CH4

=> Produces 1 mole of CO2 and 2 moles of H2O

This means that

For 0.5 mole of CH4

=> Produces 0.5 mole of CO2 and 1 mole of H20

For 2 moles of O2

=> Produces 1 mole of CO2 and 2 moles of H2O

This means that

For 1 mole of O2

=> Produces 0.5 mole of CO2 and 1 mole of H2O

This means that

For 0.5 mole of O2

=> Produces 0.25 mole of CO2 and 0.5 moles of H2O

Since, O2 produces the lesser amount of products, thus, it is the Limiting Reagent.

Therefore, CH4 will be left as it is the reagent in excess.

Thus, with every 2 moles of O2

1 mole of CH4 is used.

This means that

With every 1 mole of O2 0.5 mole of CH4 are used

With every 0.5 moles of O2 0.25 moles of CH4 are used.

This means that in the reaction, quantity of

CH4 = (0.5–0.25) moles = 0.25 moles

O2 = 0 mole

CO2 = 0.25 moles

H2O = 0.5 moles

Therefore mass of a compound

= (Moles of compound that are present) ÷ (Molecular mass of that compound)

**Mass of

CH4 = 0.25 × 16 = 4 grams

O2 = 0 × 32 = 0 gram

CO2 = 0.25 × 44 = 11 grams

H2O = 0.5 × 18 = 9 grams **

Sorry if there's any mistake. Thank you.

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We don’t have your requested question, but here is a suggested video that might help.

The equation for the complete combustion of methane is $$\mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ What volume of oxygen at SATP is needed to react exactly with $10 g$ of methane? (Section 8.2 )