Answer :
2C4H10 + 13O2 ---> 8CO2 + 10H2O
1 mole of C4H10 = 58g
According to the reaction:
2*58g of C4H10 ------------- 8 molesof CO2
58g of C4H10 ----------------- x moles of CO2
x = 4 miles od CO2
Answer: 4 moles of [tex]CO_2[/tex] are produced.
Explanation:
To calculate the number of moles, we use the formula:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
For butane:
Given mass of butane = 58g
Molar mass of butane = 58 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of butane}=\frac{58g}{58g/mol}=1mole[/tex]
For the given chemical equation:
[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]
By Stoichiometry of the reaction:
2 moles of butane produces 8 moles of carbon dioxide
So, 1 mole of butane will produce = [tex]\frac{8}{2}\times 1=4moles[/tex] of carbon dioxide
Hence, 4 moles of [tex]CO_2[/tex] are produced.
