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superman1987

Answer:

H₂O.

Explanation:

  • It is clear from the balanced equation:

CH₄ + 2H₂O → CO₂ + 4H₂.

that 1.0  mole of CH₄ reacts with 2.0 moles of H₂O to produce 1.0 mole of CO₂ and 4.0 moles of H₂.

  • To determine the limiting reactant, we should calculate the no. of moles of (20 g) CH₄ and (15 g) H₂O using the relation:

n = mass/molar mass

no. of moles of CH₄ = mass/molar mass = (20 g)/(16 g/mol) = 1.25 mol.

no. of moles of H₂O = mass/molar mass = (15 g)/(18 g/mol) = 0.833 mol.

  • from the balanced reaction, 1.0  mole of CH₄ reacts with 2.0 moles of H₂O.

So, from the calculated no. of moles: 0.4167 mole of CH₄ reacts completely with 0.833 mole of H₂O and the remaining of CH₄ will be in excess.

So, the limiting reactant is H₂O.

Eduard22sly

The limiting reactant for the reaction between 20 g of methane, CH₄ and 15 g of water, H₂O is H₂O

Balanced equation

CH₄ + 2H₂O → CO₂ + 4H₂

Molar mass of CH₄ = 12 + (1×4) = 16 g/mol

Mass of CH₄ from the balanced equation = 1 × 16 = 16 g

Molar mass of H₂O = (1×2) + 16 = 18 g/mol

Mass of H₂O from the balanced equation = 2 × 18 = 36 g

SUMMARY

From the balanced equation above,

16 g of CH₄ reacted with 36 g of H₂O

How to determine the limiting reactant

From the balanced equation above,

16 g of CH₄ reacted with 36 g of H₂O

Therefore,

20 g of CH₄ will react with = (20 × 36) / 16 = 45 g of H₂O

From the calculation made above, we can see that a higher mass of H₂O (i.e 45 g) than what was given (i.e 15 g) is needed to react completely with 20 g of CH₄.

Therefore, we can conclude that H₂O is the limiting reactant

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