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White raven [17]
3 years ago
11

What is the limiting reactant when 20g CH4 react with 15g H20?

Chemistry
1 answer:
Art [367]3 years ago
4 0

Answer:

H₂O.

Explanation:

  • It is clear from the balanced equation:

<em>CH₄ + 2H₂O → CO₂ + 4H₂.</em>

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:

<em>n = mass/molar mass</em>

<em></em>

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.

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

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.

<u><em>So, the limiting reactant is H₂O.</em></u>

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<h3>Computation for the moles of gas</h3>

The gi<em>ve</em>n gas has standard pressure, P=1\rm atm

The volume of the gas has been, V= 0. 500 \;\rm  L

The temperature of the gas has been, T=15^\circ \text C\\&#10;T=288\;\rm K

Substituting the values for the moles of gas, <em>n:</em>

<em />

<em />\rm 1\;\times\; 0. 500 =\textit n\;\times\;0.08214\;atm.L/mol.K\;\times\;288\;K\\\\&#10;\textit n=\dfrac{0. 500}{0.08214\;\times\;288} \;mol\\\\&#10;\textit n=0.021\;mol

The moles of gas in the bottle has been 0.021 mol.

Learn more about ideal gas, here:
brainly.com/question/8711877

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