#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.
Get the answer to your homework problem.
Try Numerade free for 30 days
Fordham University
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 )