Question

What mass of carbon dioxide would be produced if 100g of butane (C4H10) are completely oxidized to carbon dioxide and water?

Answer 1

By using stoichiometry and the balanced chemical equation for the combustion of butane, we determine that 302.75g of CO₂ would be produced from the complete oxidation of 100g of butane.

Step-by-step explanation:

To calculate the mass of carbon dioxide produced by the complete combustion of 100g of butane, we must first write the balanced chemical equation for the reaction:

2 C₄H₁₀(g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(g)

From the balanced equation, we see that for every 2 moles of butane, 8 moles of CO₂ are produced. Therefore, we can produce 4 moles of CO₂ for every mole of butane:

1.72 moles C₄H₁₀ × (8 moles CO₂ / 2 moles C₄H₁₀) = 6.88 moles CO₂

The molar mass of CO₂ is 44.01 g/mol. To find the mass of CO₂ produced, we multiply the number of moles by the molar mass:

6.88 moles CO₂ × 44.01 g/mol = 302.75 g CO₂

Therefore, 302.75g of CO₂ would be produced when 100g of butane is completely oxidized.

Answer 2

Answer : The mass of carbon dioxide produced will be, 303.424 grams.

Explanation : Given,

Mass of butane = 100 g

Molar mass of butane = 58 g/mole

Molar mass of carbon dioxide = 44 g/mole

The balanced chemical reaction will be,

`2C_4H_(10)+13O_2\rightarrow 8CO_2+10H_2O`

First we have to calculate the moles of
`C_4H_(10)`

`\text{Moles of }C_4H_(10)=\frac{\text{Mass of }C_4H_(10)}{\text{Molar mass of }C_4H_(10)}=(100g)/(58g/mole)=1.724moles`

Now we have to calculate the moles of
`CO_2`

From the balanced reaction, we conclude that

As, 2 moles of butane react to give 8 moles of carbon dioxide

So, 1.724 moles of butane react to give
`(8)/(2)* 1.724=6.896` moles of carbon dioxide

Now we have to calculate the mass of
`CO_2`

`\text{Mass of }CO_2=\text{Moles of }CO_2* \text{Molar mass of }CO_2`

`\text{Mass of }CO_2=(6.896mole)* (44g/mole)=303.424g`

Therefore, the mass of carbon dioxide produced will be, 303.424 grams.