13.0 mol 3.0 mol methanegasand oxygengas react to form carbon dioxide gas and watervapor. suppose you have of and of in a reactor. suppose as much as possible of the reacts. how much will be left? round your answer to the nearest

Respuesta :

The no. of moles of [tex]CH_4[/tex] remaining after the reaction is found to be 11.5 moles.

The no. of moles of a material equals the ratio of its given mass to the mass of one mole of that molecule in a chemical reaction.

The coefficients reflect the number of moles, not simply molecules, that react.

Stoichiometric coefficients are the numbers that come before the chemical formula of a molecule participating in a reaction. They assist us in determining the relative amounts or no. of moles of each reactant consumed and product created during a reaction.

[tex]CH_4 (g) + 2O_2 (g) \rightarrow CO_2 (g) + 2H_2O (g)[/tex]

No. of moles of [tex]CH_4[/tex] = 13.0 mol

No. of moles of [tex]O_2[/tex] = 3.0 mol

Here [tex]O_2[/tex] is limiting reagent because moles of [tex]O_2[/tex] is less.

Thus, [tex]CH_4[/tex] is in excess

1 mol [tex]CH_4[/tex] = 2 mol [tex]O_2[/tex]

x mol [tex]CH_4[/tex] = 3 mol [tex]O_2[/tex]

Moles of [tex]CH_4[/tex] used = 3 x 1 / 2 = 1.5 mol

Here we have 13 moles of [tex]CH_4[/tex],

Therefore,

No. of moles of [tex]CH_4[/tex] left = 13 - 1.5 = 11.5

Result:

11.5 moles of [tex]CH_4[/tex] will remain after the reaction.

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