8. Butane is burned in a lighter at a rate of 0.24 mol/min in the following reaction.
2 C4H10(g) + 13/2 O2(g) --> 4 CO2(g)+ 5H20(g)
(a) What is the rate at which O2(g) is consumed and CO2(g) is produced?
(b) How long does 15.0 g of C4H10(g) take to burn?
The equation of the combustion of butane is not balanced. Corrected is
C4H10(g) + 13/2 O2(g) --> 4 CO2(g)+ 5H2O(g) or
2C4H10(g) + 13O2(g) --> 8CO2(g)+ 10H2O(g)
a. for O2 it is 0.2 mol/min C4H10 x (13 mols O2/2 mols C4H10) = ? min
for CO2 it is 0.2 mol/min C4H10 x (8 mols CO2/2 mols C4H10) = ? min
b. You have 15 g C4H10. mols C4H10 = 15/58 = about 0.3 but you need a better number.
0.2 mols/min C4H10 x # min = approx 0.3.
Solve for # min to burn the 15 g C4H10.
To find the rate at which O2(g) is consumed and CO2(g) is produced, we need to determine the stoichiometric ratios between the reactants and the products.
(a)
From the balanced equation, we can see that for every 2 moles of C4H10, we need 13/2 moles of O2, and we produce 4 moles of CO2.
Given that the rate of C4H10 consumption is 0.24 mol/min, we can calculate the rate of O2 consumption and CO2 production as follows:
Rate of O2 consumption = (0.24 mol/min) × (13/2 mol O2 / 2 mol C4H10) = 0.78 mol/min
Rate of CO2 production = (0.24 mol/min) × (4 mol CO2 / 2 mol C4H10) = 0.48 mol/min
Therefore, the rate at which O2(g) is consumed is 0.78 mol/min, and the rate at which CO2(g) is produced is 0.48 mol/min.
(b)
To calculate how long 15.0 g of C4H10 takes to burn, we first need to convert the mass of C4H10 to moles.
The molar mass of C4H10 is:
4(12.01 g/mol) + 10(1.01 g/mol) = 58.12 g/mol
So, the number of moles of C4H10 in 15.0 g is:
15.0 g C4H10 × (1 mol C4H10 / 58.12 g C4H10) = 0.258 mol C4H10
From the balanced equation, we know that 2 moles of C4H10 are consumed per mole of O2. So, the number of moles of O2 required to burn 0.258 mol of C4H10 is:
0.258 mol C4H10 × (13/2 mol O2 / 2 mol C4H10) = 0.839 mol O2
Now, we can calculate the time it takes for 0.839 mol of O2 to be consumed:
Time = Number of moles / Rate of consumption
Time = 0.839 mol O2 / (0.24 mol/min) = 3.495 min
Therefore, it takes approximately 3.495 minutes for 15.0 g of C4H10 to burn.