Consider the following balanced equation for the combustion of butane, a fuel of

Question

Consider the following balanced equation for the combustion of butane, a fuel often used in lighters.
2C4H10(g)+13O2(g)?8CO2(g)+10H2O(g) Complete the following table, showing the appropriate masses of reactants and products. If the mass of a reactant is provided, fill in the mass of other reactants required to completely react with the given mass, as well as the mass of each product formed. If the mass of a product is provided, fill in the required masses of each reactant to make that amount of product, as well as the mass of the other product that is formed.

Mass C4H10 Mass O2 Mass CO2 Mass H2O 1.31 g 5.72g 11.12g 8.84g 222 mg 148mg

Explanation / Answer

balanced equation:--- 2C4H10(g)+13O2(g) --> 8CO2(g)+10H2O(g)

1.) moles of O2 = mass/molar mass = 1.31/32 = 0.041 moles

--> balance reaction says that 13 mole O2 react with 2 mole butane so,

for 0.041 mole O2 requires butane = 0.041 x 2/13 = 6.31 x 10^-3 mole

mass of butane = 6.31 x 10^-3 x 58 = 0.366 grams

--> moles of CO2 = 8/13 x moles of O2 = 8/13 x 0.041 = 0.02523 moles

mass of CO2 = moles x molar mass = 0.02523 x 44 = 1.11 grams

--> moles of H2O = 10/13 x 0.041 = 0.03154 moles

mass of H2O = 0.03154 x 18 = 0.568 gram

similarly all problems will be done i put here answer only....

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S.No. C4H10 O2 CO2 H2O 2.) 5.72 20.51 17.34 8.876 3.) 3.69 13.24 11.2 5.73 4.) 5.70 20.43 17.29 8.84 5.) 222 mg 796.14 mg 673.66 mg 344.48 mg 6.) 48.77 mg 174.91 mg 148 mg 75.682 mg

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