# how many grams of CO2 are produced when 88 grams of O2 are reacted with an excess of butane?

## Oh, boy! We've got some chemistry going on here! Well, to answer your question, let's break it down. The equation for the combustion of butane is:

2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

So, if we have an excess of butane, we only need to worry about the O2 being consumed. The stoichiometry of the reaction tells us that 13 moles of O2 react to produce 8 moles of CO2.

Now, to find the number of grams of CO2 produced, we need to convert grams of O2 to moles using the molar mass of O2 (32 g/mol). We divide the given mass (88 g) by the molar mass:

88 g O2 ÷ 32 g/mol = 2.75 moles of O2

Since the reaction ratio is 13 moles of O2 to 8 moles of CO2, we can set up a ratio:

2.75 moles O2 → x moles CO2

Using this ratio, we can find the number of moles of CO2:

(2.75 moles O2)(8 moles CO2 / 13 moles O2) = 1.69 moles CO2

Lastly, we convert moles of CO2 to grams using the molar mass of CO2 (44 g/mol):

1.69 moles CO2 × 44 g/mol = 74.36 grams of CO2

So, approximately 74.36 grams of CO2 are produced when 88 grams of O2 react with an excess of butane. Phew! That was a mouthful!

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## To determine the number of grams of CO2 produced when 88 grams of O2 react with an excess of butane, we need to first balance the chemical equation representing the reaction between oxygen (O2) and butane (C4H10).

The balanced equation for the combustion of butane is:

2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O

From the balanced equation, we can see that 2 moles of butane (C4H10) react with 13 moles of oxygen (O2) to produce 8 moles of carbon dioxide (CO2).

To calculate the mass of CO2 produced, we can use the molar masses of the compounds involved:

Molar mass of C4H10 = 4 x (12.01 g/mol) + 10 x (1.01 g/mol) = 58.12 g/mol
Molar mass of O2 = 2 x (16.00 g/mol) = 32.00 g/mol
Molar mass of CO2 = 12.01 g/mol + 2 x (16.00 g/mol) = 44.01 g/mol

Now, let's compute the number of moles of O2 available:

Number of moles of O2 = Mass of O2 / Molar mass of O2
= 88 g / 32.00 g/mol
= 2.75 mol

According to the balanced equation, 13 moles of O2 react with 8 moles of CO2, so the number of moles of CO2 produced will be:

Number of moles of CO2 = (Number of moles of O2 / 13) x 8
= (2.75 mol / 13) x 8
= 1.69 mol

Finally, we can calculate the mass of CO2 produced:

Mass of CO2 = Number of moles of CO2 x Molar mass of CO2
= 1.69 mol x 44.01 g/mol
= 74.24 g

Therefore, approximately 74.24 grams of CO2 are produced when 88 grams of O2 react with an excess of butane.

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## To determine the amount of CO2 produced when 88 grams of O2 react with an excess of butane, we need to know the balanced chemical equation for the combustion reaction of butane (C4H10) with oxygen (O2).

The balanced equation for the combustion of butane with oxygen is:
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

From this equation, we can see that 2 moles of butane react with 13 moles of oxygen to produce 8 moles of CO2.

Given that the molar mass of O2 is approximately 32 grams/mole, we can calculate the number of moles of O2 present in 88 grams by dividing the mass by the molar mass:
88 g O2 / 32 g/mol = 2.75 moles

Using the stoichiometry of the balanced equation, we can determine the number of moles of CO2 produced:
2.75 moles O2 × (8 moles CO2 / 13 moles O2) = 1.69 moles CO2

Finally, we can convert moles of CO2 to grams by multiplying by the molar mass of CO2, which is approximately 44 grams/mole:
1.69 moles CO2 × 44 g/mol = 74.36 grams CO2

Therefore, approximately 74.36 grams of CO2 are produced when 88 grams of O2 react with an excess of butane.

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## 2C4H10 + 13O2 ==> 8CO2 + 10H2O

mols O2 = grams/molar mass
Using the coefficients in the balanced equation, convert mols O2 to mols CO2.
Now convert mols CO2 to grams. g = mols x molar mass