How many moles of CO_2CO2 form when 58.0 g of butane, C_4H_10C4H10, burn in oxygen?

1 Answer
Meave60
Dec 1, 2016

"58.0 g butane"58.0 g butane in oxygen will produce "3.99 mol carbon dioxide"3.99 mol carbon dioxide.

Explanation:

Always start with a balanced equation.

"2C"_4"H"_10" + 13O"_2"2C4H10 + 13O2rarr→"8CO"_2" + 10H"_2"O"8CO2 + 10H2O

Determine the molar mass of butane.

Molar Mass of "C"_4"H"_10":C4H10:

(4xx12.011"g/mol")+(10xx1.008"g/mol")="58.124 g/mol"(4×12.011g/mol)+(10×1.008g/mol)=58.124 g/mol

Determine the Mole Ratios for Butane and Carbon dioxide

(2"mol C"_4"H"_10)/(8"mol CO"_2")2mol C4H108mol CO2 and (8"mol CO"_2)/(2"mol C"_4"H"_10)8mol CO22mol C4H10

Divide the given mass of butane by its molar mass (multiply by its reciprocal). This will give the moles of butane. Multiply times the molar ratio that places carbon dioxide in the numerator. This will give the moles of carbon dioxide.

58.0cancel("g C"_4"H"_10)xx(1cancel("mol C"_4"H"_10))/(58.124cancel("g C"_4"H"_10))xx(8"mol CO"_2)/(2cancel("mol C"_4"H"_10))="3.99 mol CO"_2"

Related questions
See all questions in Chemical Reactions and Equations
Impact of this question
22687 views around the world
You can reuse this answer
Creative Commons License