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aleksandr82 [10.1K]
3 years ago
13

When octane (C8H18) is burned in the presence of oxygen, the yield of products (carbon dioxide and water) is 87%. What mass of c

arbon dioxide will be produced in this engine when 21.0 g of octane is burned with 19.0 g of oxygen gas
Chemistry
1 answer:
ahrayia [7]3 years ago
6 0

Answer:

14.5g of CO₂ are produced

Explanation:

The reaction of octane with oxygen is:

C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O

<em>Where 1 mole of octane (Molar mass: 114.23g/mol) reacts with 25/2 moles of O₂ (Molar mass 32g/mol) to produce 8 moles of CO₂ and 9 moles of water.</em>

When 21.0 g of octane is burned with 19.0 g of oxygen gas you need to find <em>limiting reactant </em>to find how many moles of products are formed:

Octane: 21.0g ₓ (1mol / 114.23g) = 0.184 moles octane

Oxygen: 19.0g ₓ (1 mol / 32g) = 0.594 moles oxygen

For a complete reaction of 0.184 moles of octane you will need:

0.184 moles C₈H₁₈ ₓ (25/2 moles O₂ / 1 mole C₈H₁₈) = <em>2.3 moles of oxygen</em>

As you have just 0.594 moles of oxygen, <em>Oxygen is limiting reactant.</em>

Based on chemical equation, 25/2 of O₂ produce 8 moles of CO₂, that means theoretical yield of CO₂ with 0.594 moles of O₂ is:

0.594 moles O₂ ₓ (8 moles CO₂ / 25/2 moles O₂) = 0.380 moles of CO₂

But, as yield of products is 87%, moles produced of CO₂ are:

0.380 moles of CO₂ ₓ 87% = 0.331 moles CO₂ are produced.

As molar mass of CO₂ is 44g/mol, mass of CO₂ in 0.331 moles is:

0.331 moles CO₂ ₓ (44g / mol) =

<h3>14.5g of CO₂ are produced</h3>
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Answer:

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Explanation:

According the global reaction:

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Assuming the formula is Fe₁(CO)₅ the molecular weight is 195,9 g/mol. Thus:

1,959 Fe(CO)₅ grams  × \frac{1 mole}{195,9 Fe(CO)5 g} = 1,00×10⁻² Fe(CO)₅ moles

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1 Fe(CO)₅ + ¹³/₂O₂ → ¹/₂ Fe₂O₃ + 5 CO₂

With this balanced equation the moles produced have sense, thus, the empirical formula is: Fe(CO)₅

I hope it helps!

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2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.

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Explanation:

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So the molar mass of CaF₂ is:

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Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 1000 grams of the compound how many moles are there?

moles=\frac{1000 grams*1 mole}{78 grams}

moles=12.82

<u><em>12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution</em></u>

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