Question

If 20 grams of butane, C4H10, is burned in a combustion reaction, how many grams of carbon dioxide could you get?

Answer 1

To calculate the production of CO2 from burning butane, use stoichiometry based on the balanced equation. Approximately 60.57 grams of CO2 can be produced from 20 grams of butane.

Step-by-step explanation:

To calculate the mass of carbon dioxide produced from the combustion of butane (C4H10), we first need the balanced chemical equation:

2 C4H10 + 13 O2 → 8 CO2 + 10 H2O.

This equation indicates that from every 2 moles of butane combusted, 8 moles of carbon dioxide are produced. Utilizing the molar mass of butane (58.12 g/mol) and carbon dioxide (44.01 g/mol), we can find how many grams of carbon dioxide can be produced from 20 grams of butane through stoichiometry.

Firstly, calculate the number of moles of butane in 20 grams:

Moles of butane = mass / molar mass = 20 g / 58.12 g/mol ≈ 0.344 moles of butane

According to the balanced equation, 1 mole of butane would produce 4 moles of CO2. Therefore, 0.344 moles of butane would produce:

0.344 moles of butane x (4 moles of CO2 / 1 mole of butane) ≈ 1.376 moles of CO2

Now, convert the moles of carbon dioxide to grams:

Grams of CO2 = moles of CO2 x molar mass of CO2 = 1.376 moles x 44.01 g/mol ≈ 60.57 grams of CO2

Thus, you could get approximately 60.57 grams of carbon dioxide from the combustion of 20 grams of butane.

Answer 2

Step 1 - Balancing the combustion equation

The combustion of an alkane always gives, as products, H2O and CO2. Remembering that combustion always involves a reaction with O2, we can write the chemical equation for the combustion of C4H10 as:

`C_4H_(10(g))+(13)/(2)O_(2(g))\rightleftarrows4CO_(2(g))+5H_2O`

Step 2 - Discover how many grams of carbon dioxide can be produced

Looking at the equation above, we can see that 1 mole of C4H10 produces 4 moles of CO2. This is a fixed proportion. We can convert this proportion in moles to a proportion in mass by multiplying each number of moles by the respective molar mass (C4H10 58g/mol; CO2 44g/mol)of the substances:

`\begin{gathered} mC_4H_(10)=1*58=58\text{ g} \\ \\ mCO_2=4*44=176\text{ g} \end{gathered}`

We discovered thus that each 58 g of C4H10 produce 176g of CO2. Since this is a fixed proportion in mass, we can use it to discover how many grams of CO2 would be formed by the combustion of 20g of butane (C4H10):

`\begin{gathered} 58\text{ g of C4H10------ 176 g of CO2} \\ 20\text{ g of C4H10 ----- x} \\ \\ x=(176*20)/(58)=60.7\text{ g of CO2} \end{gathered}`

The combustion of 20 g of C4H10 would produce thus 60.7 g of CO2.