Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
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:
[tex]C_4H_{10(g)}+\frac{13}{2}O_{2(g)}\rightleftarrows4CO_{2(g)}+5H_2O[/tex]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:
[tex]\begin{gathered} mC_4H_{10}=1\times58=58\text{ g} \\ \\ mCO_2=4\times44=176\text{ g} \end{gathered}[/tex]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):
[tex]\begin{gathered} 58\text{ g of C4H10------ 176 g of CO2} \\ 20\text{ g of C4H10 ----- x} \\ \\ x=\frac{176\times20}{58}=60.7\text{ g of CO2} \end{gathered}[/tex]The combustion of 20 g of C4H10 would produce thus 60.7 g of CO2.
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
