How many moles of #CO_2# form when 58.0 g of butane, #C_4H_10#, burn in oxygen?

Question

Natalie Anton · Accepted Answer

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

Always start with a balanced equation. #"2C"_4"H"_10" + 13O"_2"##rarr##"8CO"_2" + 10H"_2"O"# Determine the molar mass of butane. Molar Mass of #"C"_4"H"_10":# #(4xx12.011"g/mol")+(10xx1.008"g/mol")="58.124 g/mol"# Determine the Mole Ratios for Butane and Carbon dioxide #(2"mol C"_4"H"_10)/(8"mol CO"_2")# and #(8"mol CO"_2)/(2"mol C"_4"H"_10)# 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"#

Ellie Andrews · Answer

undefined The chemical equation for the combustion of butane, 2C4H10 + 13O2 -> 8CO2 + 10H2O, must first be balanced in order to determine how many moles of butane are produced when 58.0 g of butane burns in oxygen. Based on this equation, we can see that 2 moles of butane produce 8 moles of CO2. Next, we calculate the number of moles of butane: molar mass of butane (C4H10) = 4(12.01) + 10(1.01) = 58.12 g/mol; moles of butane = mass / molar mass = 58.0 g / 58.12 g/mol ≈ 1.00 mol. Since 2 moles of butane produce 8 moles of CO2, 1 mole of butane will produce 8/2 = 4 moles of CO2.