Respuesta :
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.
