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4vir4ik [10]
2 years ago
13

Methane (CH4), is said to be the cleanest burning fossil fuel. When burned, it releases carbon dioxide gas and forms water. Usin

g the equation below, determine which of the following is true (assuming there is an infinite supply of oxygen gas (O2)). Select all that apply.
For every 1 molecule of methane CH4 that reacts, 2 molecules of H2O are produced.


For every 2 moles of methane (CH4) that reacts, 2 moles of H2O are produced.


For every 2,000 molecules of methane (CH4) that reacts, 2,000 molecules of H2O are produced.


For every 20 grams of methane (CH4) that reacts, 40 grams of H2O are produced.


For every 200 moles of methane (CH4) that reacts, 400 moles of H2O are produced.
Chemistry
1 answer:
LiRa [457]2 years ago
3 0

From the stoichiometry of the balanced reaction equation, the correct statement are;

  • For every 1 molecule of methane CH4 that reacts, 2 molecules of H2O are produced.
  • For every 20 grams of methane (CH4) that reacts, 40 grams of H2O are produced.
  • For every 200 moles of methane (CH4) that reacts, 400 moles of H2O are produced.

<h3>What is combustion?</h3>

The term combustion refers to the burning of fossil fuels for the purpose of energy production. The equation for reaction is CH4 + 2O2 ---> CO2 + 2H2O.

Using this equation as shown, the true statements are;

  • For every 1 molecule of methane CH4 that reacts, 2 molecules of H2O are produced.
  • For every 20 grams of methane (CH4) that reacts, 40 grams of H2O are produced.
  • For every 200 moles of methane (CH4) that reacts, 400 moles of H2O are produced.

Learn more about combustion: brainly.com/question/15117038

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First we have to calculate the moles of NaNO_3.

\text{Moles of }NaNO_3=\frac{\text{Mass of }NaNO_3}{\text{Molar mass of }NaNO_3}=\frac{244g}{85g/mole}=2.87moles

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The balanced chemical reaction is,

Pb(NO_3)_2(aq)+2NaBr(aq)\rightarrow PbBr_2(s)+2NaNO_3(aq)

From the balanced reaction we conclude that

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So, 2.87 moles of NaNO_3 react with 2.87 moles of NaBr

Now we have to calculate the mass of NaBr.

\text{Mass of }NaBr=\text{Moles of }NaBr\times \text{Molar mass of }NaBr

\text{Mass of }NaBr=(2.87mole)\times (102.9g/mole)=295.323g

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