what mass of CO2 can be formed when 14.17g of C4H10 are burned in excess oxygen
I do not even know where to start
4*12+10 = 58 grams/mol
14.17/58 = .244 mol
2C4H10 + 13 O2 ---> 10H2O + 8 CO2
mols CO2 produced = .244(8/2)=.977 mol
2*12+16 = 40 grams/mol
40 g/mol * .977 mol = 39.1 grams
To find the mass of CO2 formed when 14.17g of C4H10 is burned, we need to determine the balanced chemical equation for the combustion reaction of C4H10 (butane) with oxygen (O2).
The balanced chemical equation for the combustion of butane is:
2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O
From the equation, we can see that for every 2 moles of C4H10 burned, 8 moles of CO2 are formed.
To begin, we need to convert the given mass of C4H10 (butane) to moles. You can do this by using the molar mass of C4H10, which is 58.12 g/mol.
moles of C4H10 = mass of C4H10 / molar mass of C4H10
moles of C4H10 = 14.17g / 58.12 g/mol
Next, we use the stoichiometry of the balanced equation to find the moles of CO2 formed from the moles of C4H10. From the balanced equation, we know that for every 2 moles of C4H10, 8 moles of CO2 are produced.
moles of CO2 = moles of C4H10 * (8 moles of CO2 / 2 moles of C4H10)
Finally, we convert the moles of CO2 to grams by multiplying it by the molar mass of CO2, which is 44.01 g/mol.
mass of CO2 = moles of CO2 * molar mass of CO2
By substituting the values, you can calculate the mass of CO2 formed.